If the equation 3x^2 + (3 - p)xy + qy^2 - 2px | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to represent a circle, the coefficient of $xy$ must be $0$ and the coefficients of $x^2$ an

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) For the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to represent a circle, the coefficient of $xy$ must be $0$ and the coefficients of $x^2$ and $y^2$ must be equal.Step $1$: Set the coefficient of $xy$ to $0$: $3 - p = 0 \implies p = 3$.Step $2$: Set the coefficients of $x^2$ and $y^2$ equal: $3 = q$.Step $3$: Substitute $p=3$ and $q=3$ into the equation: $3x^2 + 3y^2 - 2(3)x = 8(3)(3) \implies 3x^2 + 3y^2 - 6x = 72$.Step $4$: Divide by $3$: $x^2 + y^2 - 2x = 24$.Step $5$: Complete the square: $(x^2 - 2x + 1) + y^2 = 24 + 1 \implies (x - 1)^2 + y^2 = 25$.Step $6$: The radius $r$ is $\sqrt{25} = 5$. The area is $\pi r^2 = \pi(5)^2 = 25\pi$ sq. units.

If the equation $3x^2 + (3 - p)xy + qy^2 - 2px = 8pq$ represents a circle, then the area (in sq. units) of this circle is (in $\pi$)

Similar Questions

The Cartesian equation of the curve given by $x = 6 \cos \theta$ and $y = 6 \sin \theta$ is

Find the equation of the circle with radius $5$ whose centre lies on the $x-$axis and passes through the point $(2, 3)$.

Find the equation of a circle whose diameters are $2x - 3y = 5$ and $3x - 4y = 7$ and whose radius is $8$.

The lengths of the intercepts made by a circle $S$ on $X$ and $Y$-axes are $\frac{2 \sqrt{13}}{3}$ and $\frac{2 \sqrt{22}}{3}$ respectively. If the radius of the circle $S$ is $\frac{\sqrt{38}}{3}$ and its centre $C$ lies in the second quadrant,then $C=$

Let the equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a point circle other than the origin. Then which one of the following conditions must hold?

Vedclass Test Series

Exam Paper Generator

Online Exam Module