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 radius of the circle $r=12 \cos \theta+5 \sin \theta$ is

The centre and radius of a circle $x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right), y=\frac{8 a t}{1+t^{2}}$ are respectively:

The equation of the circle passing through $(4, 5)$ and having the centre at $(2, 2)$ is:

The equation of the circle whose diameters have the end points $(a, 0)$ and $(0, b)$ is given by:

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units. The equation of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module