Questions and Answers Section

Example 1: Finding Center and Radius of a Circle

Question: Find the center and radius of the circle $x^2+y^2-8x+10y-12=0$.

Solution: To solve this, we compare the given equation with the general form of a circle: $x^2+y^2+2gx+2fy+c=0$. 🔵

Step 1: Identify g, f, and c.

• Comparing the x-terms: $2gx = -8x \implies g = -4$.
• Comparing the y-terms: $2fy = 10y \implies f = 5$.
• Comparing the constant terms: $c = -12$.

Step 2: Find the center. The center of the circle is given by the formula $(-g, -f)$. Center = $(-(-4), -5) = \mathbf{(4, -5)}$.

Step 3: Find the radius. The radius is given by the formula $r = \sqrt{g^2+f^2-c}$. $r = \sqrt{(-4)^2+5^2-(-12)} = \sqrt{16+25+12} = \mathbf{\sqrt{53}}$.

Example 2: Equation of a Circle in Diameter Form

Question: Find the equation of the circle whose diameter has endpoints A(-2,3) and B(4,5).

Solution: We use the diameter form of the circle's equation, which is a shortcut based on the property that any angle inscribed in a semicircle is a right angle.

The formula is: $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.

Step 1: Substitute the endpoint coordinates. Here, $(x_1, y_1) = (-2,3)$ and $(x_2, y_2) = (4,5)$. $(x - (-2))(x - 4) + (y - 3)(y - 5) = 0$ $(x+2)(x-4) + (y-3)(y-5) = 0$

Step 2: Expand the terms. $(x^2 - 4x + 2x - 8) + (y^2 - 5y - 3y + 15) = 0$ $(x^2 - 2x - 8) + (y^2 - 8y + 15) = 0$

Step 3: Combine and write the final equation. $x^2+y^2-2x-8y+7=0$.

Example 3: Circle Touching an Axis

Question: Find the equation of the circle that is concentric with $x^2+y^2-4x-6y-3=0$ and touches the y-axis.

Solution: Step 1: Find the center of the given circle. 'Concentric' means they share the same center. For $x^2+y^2-4x-6y-3=0$, we have $g=-2$ and $f=-3$. The center is $(-g, -f) = \mathbf{(2, 3)}$.

Step 2: Determine the radius of the new circle. The new circle also has its center at (2, 3). For a circle to touch the y-axis, its radius must be equal to the horizontal distance from its center to the y-axis. This distance is the absolute value of the x-coordinate of its center. So, $r = |2| = \mathbf{2}$.

Step 3: Write the equation. Using the center-radius form $(x-h)^2+(y-k)^2=r^2$ with center (2,3) and radius 2: $(x-2)^2+(y-3)^2=2^2$ $(x-2)^2+(y-3)^2=4$.

Example 4: Orthogonal Circles

Question: Find the value of k if the circles $x^2+y^2+4x+8=0$ and $x^2+y^2-16y+k=0$ cut orthogonally.

Solution: Two circles cut orthogonally (at a 90° angle) if the tangents at their points of intersection are perpendicular. This leads to the condition: $2g_1g_2+2f_1f_2=c_1+c_2$. 📐

Step 1: Identify the parameters for both circles.

• Circle 1: $x^2+y^2+4x+8=0$ $2g_1=4 \implies g_1=2$, $2f_1=0 \implies f_1=0$, $c_1=8$.
• Circle 2: $x^2+y^2-16y+k=0$ $2g_2=0 \implies g_2=0$, $2f_2=-16 \implies f_2=-8$, $c_2=k$.

Step 2: Apply the orthogonality condition. $2(2)(0)+2(0)(-8)=8+k$ $0 + 0 = 8+k$ $0 = 8+k$

Step 3: Solve for k. $\mathbf{k=-8}$.

Example 5: Finding Parameters of a Parabola

Question: For the parabola $y^2 = 16x$, find the focus, the equation of the directrix, and the length of the latus rectum.

Solution: Step 1: Compare with the standard form. The given equation is of the form $y^2 = 4ax$. Comparing this with $y^2 = 16x$, we have: $4a = 16 \implies \mathbf{a=4}$.

Step 2: Determine the parameters.

• Focus: The focus is at $(a,0)$. So, the focus is (4,0).
• Directrix: The equation of the directrix is $x=-a$. So, the equation is $x=-4$.
• Length of Latus Rectum: The length is given by $4a$. So, the length is 16.

Example 6: Equation of a Parabola from Parameters

Question: Find the equation of the parabola with focus at (0,-3) and directrix $y=3$.

Solution: Step 1: Determine the orientation and vertex.

• The focus S(0,-3) is on the y-axis.
• The directrix $y=3$ is a horizontal line.
• Since the focus is below the directrix, the parabola must open downwards.
• The vertex is the midpoint of the focus and the directrix along the axis of symmetry (the y-axis). The vertex is $(\frac{0+0}{2}, \frac{-3+3}{2}) = (0,0)$.

Step 2: Find the value of 'a'. The distance from the vertex (0,0) to the focus (0,-3) is $\mathbf{a=3}$.

Step 3: Write the equation. The standard equation for a downward-opening parabola with its vertex at the origin is $x^2=-4ay$. Substituting $a=3$, we get: $x^2=-4(3)y \implies \mathbf{x^2 = -12y}$.

Example 7: Parabola with a Shifted Vertex

Question: Find the equation of the parabola with vertex at (2,1) and focus at (2,4).

Solution: Step 1: Determine the orientation. The vertex is V(2,1) and the focus is F(2,4). Since the x-coordinates are the same, the axis of symmetry is the vertical line $x=2$. As the focus is above the vertex, the parabola opens upwards.

Step 2: Find 'a'. 'a' is the distance between the vertex and the focus. $a = \sqrt{(2-2)^2 + (4-1)^2} = \sqrt{0^2 + 3^2} = 3$.

Step 3: Write the equation. The standard form for an upward-opening parabola with a shifted vertex $(h,k)$ is $(x-h)^2 = 4a(y-k)$. Here, $(h,k)=(2,1)$ and $a=3$. The equation is: $(x-2)^2 = 4(3)(y-1) \implies \mathbf{(x-2)^2 = 12(y-1)}$.

Example 8: Focal Distance of a Parabola

Question: The focal distance of a point on the parabola $y^2=12x$ is 6. Find the coordinates of the point.

Solution: The focal distance of a point P(x,y) on a parabola is its distance from the focus. By definition, this is equal to its perpendicular distance from the directrix.

Step 1: Find 'a'. Comparing $y^2=12x$ with $y^2=4ax$, we get $4a=12 \implies \mathbf{a=3}$.

Step 2: Use the focal distance formula. For a right-handed parabola $y^2=4ax$, the focal distance of any point (x,y) is given by the simple formula: Distance = $x+a$. We are given that this distance is 6. $x+3=6 \implies \mathbf{x=3}$.

Step 3: Find the y-coordinate. Substitute $x=3$ into the parabola's equation: $y^2 = 12(3) = 36 \implies y = \pm \sqrt{36} = \pm 6$. The points are (3, 6) and (3, -6).

Example 9: Finding Parameters of an Ellipse

Question: Find the eccentricity and foci of the ellipse $16x^2+25y^2=400$.

Solution: Step 1: Convert to standard form. Divide the entire equation by 400: $\frac{16x^2}{400} + \frac{25y^2}{400} = 1 \implies \frac{x^2}{25} + \frac{y^2}{16} = 1$.

Step 2: Identify parameters. Since $25 > 16$, this is a horizontal ellipse with $a^2=25$ and $b^2=16$.

Step 3: Find the eccentricity (e). Using the relation $b^2=a^2(1-e^2)$: $16 = 25(1-e^2) \implies \frac{16}{25} = 1-e^2 \implies e^2 = 1 - \frac{16}{25} = \frac{9}{25}$. So, $\mathbf{e=3/5}$.

Step 4: Find the foci. The foci are at $(\pm ae, 0)$. Foci = $(\pm 5(\frac{3}{5}), 0) = \mathbf{(\pm 3, 0)}$.

Example 10: Equation of an Ellipse from Parameters

Question: Find the equation of the ellipse whose foci are at $(\pm 5, 0)$ and vertices are at $(\pm 13, 0)$.

Solution: Step 1: Determine a and c.

• The vertices are at $(\pm a, 0)$, so $\mathbf{a=13}$.
• The foci are at $(\pm c, 0)$, so $\mathbf{c=5}$.

Step 2: Find $b^2$. For an ellipse, the key relationship is $c^2=a^2-b^2$. $5^2 = 13^2 - b^2 \implies 25 = 169 - b^2$. $b^2 = 169 - 25 = \mathbf{144}$.

Step 3: Write the equation. The equation for a horizontal ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Substituting the values: $\mathbf{\frac{x^2}{169} + \frac{y^2}{144} = 1}$.

Example 11: Latus Rectum of an Ellipse

Question: Find the length of the latus rectum of the ellipse $3x^2+4y^2=12$.

Solution: Step 1: Convert to standard form. Divide the equation by 12: $\frac{3x^2}{12} + \frac{4y^2}{12} = 1 \implies \frac{x^2}{4} + \frac{y^2}{3} = 1$.

Step 2: Identify a and $b^2$. For an ellipse, $a^2$ is always the larger denominator. Since $4 > 3$, this is a horizontal ellipse. $a^2=4 \implies \mathbf{a=2}$. $b^2=\mathbf{3}$.

Step 3: Calculate the length of the latus rectum. The formula is $\frac{2b^2}{a}$. Length = $\frac{2(3)}{2} = \mathbf{3}$.

Example 12: Sum of Focal Distances of an Ellipse

Question: What is the sum of the focal distances of any point on the ellipse $9x^2+16y^2=144$?

Solution: By the definition of an ellipse, the sum of the distances of any point on the curve from its two foci is constant and is equal to the length of the major axis (2a).

Step 1: Convert the equation to standard form. Divide by 144: $\frac{9x^2}{144} + \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} + \frac{y^2}{9} = 1$.

Step 2: Find the value of 'a'. Here, $a^2=16$, so $\mathbf{a=4}$.

Step 3: Calculate the length of the major axis. The sum of the focal distances = $2a = 2(4) = \mathbf{8}$.

Example 13: Finding Parameters of a Hyperbola

Question: For the hyperbola $9x^2 - 16y^2 = 144$, find the eccentricity and foci.

Solution: Step 1: Convert to standard form. Divide by 144: $\frac{9x^2}{144} - \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} - \frac{y^2}{9} = 1$.

Step 2: Identify parameters. This is a horizontal hyperbola with $a^2=16$ and $b^2=9$.

Step 3: Find the eccentricity (e). Using the relation for a hyperbola, $b^2=a^2(e^2-1)$: $9 = 16(e^2-1) \implies \frac{9}{16} = e^2-1 \implies e^2 = 1 + \frac{9}{16} = \frac{25}{16}$. So, $\mathbf{e=5/4}$.

Step 4: Find the foci. The foci are at $(\pm ae, 0)$. Foci = $(\pm 4(\frac{5}{4}), 0) = \mathbf{(\pm 5, 0)}$.

Example 14: Equation of a Hyperbola from Parameters

Question: Find the equation of the hyperbola whose foci are $(\pm 5, 0)$ and the length of the transverse axis is 8.

Solution: Step 1: Determine a and c.

• The length of the transverse axis is $2a=8 \implies \mathbf{a=4}$.
• The foci are at $(\pm c, 0)$, so $\mathbf{c=5}$.

Step 2: Find $b^2$. For a hyperbola, the key relationship is $c^2=a^2+b^2$. $5^2 = 4^2 + b^2 \implies 25 = 16 + b^2$. $b^2 = 25 - 16 = \mathbf{9}$.

Step 3: Write the equation. Since the foci are on the x-axis, it's a horizontal hyperbola with the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substituting the values: $\mathbf{\frac{x^2}{16} - \frac{y^2}{9} = 1}$.

Example 15: Asymptotes of a Hyperbola

Question: Find the equation of the asymptotes of the hyperbola $3x^2-y^2=3$.

Solution: Step 1: Write the equation in standard form. Divide by 3: $\frac{3x^2}{3} - \frac{y^2}{3} = 1 \implies \frac{x^2}{1} - \frac{y^2}{3} = 1$.

Step 2: Identify a and b. Here $a^2=1 \implies a=1$, and $b^2=3 \implies b=\sqrt{3}$.

Step 3: Use the asymptote formula. The equations of the asymptotes for a horizontal hyperbola are $y = \pm \frac{b}{a}x$. Substituting the values gives: $\mathbf{y = \pm \sqrt{3}x}$.

Example 16: Rectangular Hyperbola

Question: Find the length of the transverse axis of the rectangular hyperbola $xy=18$.

Solution: Step 1: Understand the form. The equation $xy=c^2$ represents a rectangular hyperbola whose asymptotes are the coordinate axes. This is a rotation of the standard hyperbola $x^2-y^2=a^2$.

Step 2: Use the relationship between the forms. The relationship between the constants is $\mathbf{c^2 = a^2/2}$. Here we are given $c^2=18$. So, $18 = a^2/2 \implies a^2=36 \implies \mathbf{a=6}$.

Step 3: Find the length of the transverse axis. The length of the transverse axis is $2a = 2(6) = \mathbf{12}$.

Example 17: Tangent to a Circle

Question: Find the equation of the tangent to the circle $x^2+y^2=25$ at the point (3,-4).

Solution: The equation of the tangent to the circle $x^2+y^2=r^2$ at a point $(x_1, y_1)$ on the circle is given by the formula $xx_1+yy_1=r^2$.

Step 1: Identify the parameters.

• The point of tangency is $(x_1, y_1) = (3, -4)$.
• The circle's radius squared is $r^2=25$.

Step 2: Substitute into the formula. $x(3) + y(-4) = 25$

Step 3: Simplify. $\mathbf{3x-4y=25}$.

Example 18: Tangent to a Parabola

Question: Find the equation of the tangent to the parabola $y^2=16x$ at the point where the parameter $t=2$.

Solution: Step 1: Find 'a' and the point of tangency. For $y^2=16x$, we have $4a=16 \implies \mathbf{a=4}$. The parametric coordinates are $(at^2, 2at)$. For $t=2$, the point is: $(4(2^2), 2(4)(2)) = \mathbf{(16, 16)}$.

Step 2: Use the parametric tangent formula. The equation of the tangent to $y^2=4ax$ at the point with parameter 't' is $ty=x+at^2$. Substituting $t=2$ and $a=4$: $2y = x + 4(2^2)$ $2y = x + 16$ or $\mathbf{x-2y+16=0}$.

Example 19: Normal to an Ellipse

Question: Find the equation of the normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at the point $(4,0)$.

Solution: Step 1: Analyze the point. The point (4,0) is a vertex of the ellipse, since $a^2=16 \implies a=4$, and the vertices are at $(\pm a, 0)$.

Step 2: Determine the tangent at this point. The tangent to an ellipse at a vertex is a vertical line. In this case, the tangent is the line $\mathbf{x=4}$.

Step 3: Determine the normal. The normal is perpendicular to the tangent and passes through the same point (4,0). A line perpendicular to a vertical line is a horizontal line. The horizontal line passing through (4,0) is the x-axis. Therefore, the equation of the normal is $\mathbf{y=0}$.

Example 20: Condition of Tangency for Hyperbola

Question: Find the values of m for which the line $y=mx+2$ is a tangent to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$.

Solution: The condition for a line $y=mx+c$ to be tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $c^2=a^2m^2-b^2$.

Step 1: Identify the parameters.

• From the line, $c=2$.
• From the hyperbola, $a^2=9$ and $b^2=4$.

Step 2: Substitute into the condition and solve for m. $2^2 = 9m^2 - 4$ $4 = 9m^2 - 4$ $8 = 9m^2$ $m^2 = 8/9$ $m = \pm \sqrt{8/9} = \mathbf{\pm \frac{2\sqrt{2}}{3}}$.

Example 21: Common Chord

Question: Find the length of the common chord of the parabola $y^2=4x$ and the circle $x^2+y^2=5$.

Solution: Step 1: Find the points of intersection. Substitute $y^2=4x$ from the parabola's equation into the circle's equation: $x^2+(4x)=5 \implies x^2+4x-5=0$.

Step 2: Solve the quadratic for x. $(x+5)(x-1)=0$. This gives $x=-5$ or $x=1$. Since the parabola is $y^2=4x$, x cannot be negative for a real y. So we must have $\mathbf{x=1}$.

Step 3: Find the y-coordinates. When $x=1$, $y^2=4(1)=4 \implies y=\pm 2$. The points of intersection are (1,2) and (1,-2).

Step 4: Calculate the distance. The length of the chord is the distance between these two points. Since the x-coordinates are the same, it is a vertical line segment. Length = $|2 - (-2)| = \mathbf{4}$.

Example 22: Identifying a Conic

Question: Identify the conic represented by $4x^2+y^2-8x-2y+1=0$.

Solution: We can identify the conic by converting the equation to its standard (vertex/center) form by completing the square.

Step 1: Group x and y terms. $(4x^2-8x) + (y^2-2y) + 1 = 0$ $4(x^2-2x) + (y^2-2y) + 1 = 0$

Step 2: Complete the square for both variables. $4((x-1)^2-1) + ((y-1)^2-1) + 1 = 0$ $4(x-1)^2 - 4 + (y-1)^2 - 1 + 1 = 0$ $4(x-1)^2 + (y-1)^2 = 4$

Step 3: Convert to standard form and identify. Divide by 4: $\frac{(x-1)^2}{1} + \frac{(y-1)^2}{4} = 1$. This equation is of the form $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$. Since the coefficients of the squared terms are different and positive, and they are added together, the conic is an ellipse.

Example 23: Locus of Midpoints

Question: Find the locus of the midpoints of the focal chords of the parabola $y^2=4ax$.

Solution: Step 1: Use the equation of a chord with a given midpoint. Let the midpoint be $(h,k)$. The equation of the chord of the parabola $y^2=4ax$ with midpoint $(h,k)$ is given by the formula $T=S_1$. $T: ky-2a(x+h) = 0$ $S_1: k^2-4ah = 0$ So, the equation is $ky-2a(x+h) = k^2-4ah$.

Step 2: Apply the condition that the chord is focal. A focal chord is a chord that passes through the focus, S(a,0). We substitute these coordinates into our chord equation: $k(0) - 2a(a+h) = k^2 - 4ah$ $-2a^2 - 2ah = k^2 - 4ah$

Step 3: Simplify and find the locus. $k^2 = -2a^2 - 2ah + 4ah$ $k^2 = 2ah - 2a^2$ $k^2 = 2a(h-a)$. To get the locus, we replace the specific point $(h,k)$ with the general point $(x,y)$: $\mathbf{y^2=2a(x-a)}$.

Example 24: Parametric Point of Ellipse

Question: Find the point on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ corresponding to the eccentric angle $\theta = \pi/4$.

Solution: Step 1: Identify 'a' and 'b'. From the equation, $a^2=25 \implies a=5$ and $b^2=9 \implies b=3$.

Step 2: Use the parametric coordinate formulas. The parametric coordinates of a point on this ellipse are given by $(x,y) = (a\cos\theta, b\sin\theta)$.

Step 3: Substitute the given angle. We are given $\theta = \pi/4$. We know that $\cos(\pi/4) = 1/\sqrt{2}$ and $\sin(\pi/4) = 1/\sqrt{2}$. x-coordinate: $x = 5 \cos(\pi/4) = 5(1/\sqrt{2}) = 5/\sqrt{2}$. y-coordinate: $y = 3 \sin(\pi/4) = 3(1/\sqrt{2}) = 3/\sqrt{2}$. The point is $(5/\sqrt{2}, 3/\sqrt{2})$.

Example 25: Common Tangents

Question: Find the number of common tangents to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$.

Solution: Step 1: Find the center and radius of each circle.

• Circle 1: $C_1 = (2,3)$, $r_1=\sqrt{4+9-(-12)}=\sqrt{25}=5$.
• Circle 2: $C_2 = (-3,-9)$, $r_2=\sqrt{9+81-26}=\sqrt{64}=8$.

Step 2: Find the distance between the centers (d). $d = \sqrt{(2-(-3))^2+(3-(-9))^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$.

Step 3: Compare 'd' with the sum of the radii. Sum of radii: $r_1+r_2=5+8=13$.

Step 4: Determine the number of tangents. Since the distance between the centers is exactly equal to the sum of their radii ($d=r_1+r_2$), the circles touch each other externally. For circles in this configuration, there are 3 common tangents. ✅

Example 26: Director Circle

Question: Find the equation of the director circle of the ellipse $9x^2+16y^2=144$.

Solution: The director circle is the locus of the point of intersection of perpendicular tangents to a conic.

Step 1: Find $a^2$ and $b^2$ of the ellipse. Divide by 144: $\frac{9x^2}{144}+\frac{16y^2}{144}=1 \implies \frac{x^2}{16}+\frac{y^2}{9}=1$. So, $a^2=16$ and $b^2=9$.

Step 2: Use the formula for the director circle. For an ellipse, the equation of the director circle is $x^2+y^2=a^2+b^2$. Substituting the values: $x^2+y^2=16+9$ $x^2+y^2=25$.

Example 27: Asymptotes and Eccentricity

Question: If the angle between the asymptotes of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2\theta$, show that the eccentricity is $e=\sec\theta$.

Solution: Step 1: Relate the angle to the slope of the asymptotes. The asymptotes are $y=(b/a)x$ and $y=-(b/a)x$. The angle one of the asymptotes makes with the positive x-axis is $\theta$. The total angle between them is $2\theta$. Therefore, the slope of the asymptote $y=(b/a)x$ is $\tan\theta$. So, $\tan\theta = \frac{b}{a}$.

Step 2: Use the eccentricity formula for a hyperbola. We know that $e^2 = 1 + \frac{b^2}{a^2}$.

Step 3: Substitute and simplify. $e^2 = 1 + (\frac{b}{a})^2 = 1 + (\tan\theta)^2 = 1 + \tan^2\theta$. From the trigonometric identity $1+\tan^2\theta = \sec^2\theta$, we have: $e^2 = \sec^2\theta$. Taking the square root (and since e > 1 and $\theta$ is acute), we get $\mathbf{e=\sec\theta}$.

Example 28: Equation of Parabola with a point

Question: Find the equation of the parabola with vertex at the origin, axis along the x-axis, and passing through the point (2,3).

Solution: Step 1: Determine the standard form. Since the vertex is at the origin and the axis is the x-axis, the equation is of the form $y^2=4ax$ (opens right) or $y^2=-4ax$ (opens left). Since the point (2,3) has a positive x-coordinate, the parabola must open to the right. The form is $\mathbf{y^2=4ax}$.

Step 2: Use the given point to find 'a'. The parabola passes through (2,3), so these coordinates must satisfy the equation: $3^2 = 4a(2) \implies 9=8a \implies \mathbf{a=9/8}$.

Step 3: Write the final equation. Substitute the value of 'a' back into the standard form: $y^2 = 4(9/8)x = \mathbf{\frac{9}{2}x}$.

Example 29: Intercepts of a Circle

Question: Find the length of the y-intercept of the circle $x^2+y^2-3x-4y=0$.

Solution: Step 1: Find the points of intersection with the y-axis. To find the y-intercept, we set $\mathbf{x=0}$ in the circle's equation: $0^2+y^2-3(0)-4y=0$ $y^2-4y=0$.

Step 2: Solve for y. $y(y-4)=0$. The intersection points are at $\mathbf{y=0}$ and $\mathbf{y=4}$. These correspond to the points (0,0) and (0,4) on the circle.

Step 3: Calculate the length. The length of the y-intercept is the distance between these two points. Length = $|4 - 0| = \mathbf{4}$.

Example 30: Auxiliary Circle

Question: Find the equation of the auxiliary circle of the ellipse $4x^2+9y^2=36$.

Solution: The auxiliary circle of an ellipse is the circle that has the major axis of the ellipse as its diameter.

Step 1: Convert the ellipse equation to standard form. Divide by 36: $\frac{4x^2}{36}+\frac{9y^2}{36}=1 \implies \frac{x^2}{9}+\frac{y^2}{4}=1$.

Step 2: Find the semi-major axis 'a'. This is a horizontal ellipse with $a^2=9$, so $\mathbf{a=3}$. The major axis has length $2a=6$.

Step 3: Determine the properties of the auxiliary circle.

• Its center is the same as the ellipse's center: (0,0).
• Its radius is equal to the semi-major axis of the ellipse: $r=a=3$.

Step 4: Write the equation. The equation of a circle with center (0,0) and radius 3 is $x^2+y^2=r^2$. So, the equation is $\mathbf{x^2+y^2=9}$.