If A and B are the points of contact of the t | vedclass

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

Question

(C) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center is $C(2, -1)$ and the radius is $r = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$.

Vedclass · Accepted Answer

$x^2+y^2+x-7=0$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center is $C(2, -1)$ and the radius is $r = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$. Let $P = (-3, 1)$. The circle with diameter $PC$ has the equation $(x - (-3))(x - 2) + (y - 1)(y - (-1)) = 0$. This simplifies to $(x+3)(x-2) + (y-1)(y+1) = 0$,which is $x^2+x-6 + y^2-1 = 0$,or $x^2+y^2+x-7=0$. The points of contact $A$ and $B$ lie on this circle because $\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$. Thus,the circumcircle of $	riangle PAB$ is the circle with diameter $PC$,which is $x^2+y^2+x-7=0$.

If $A$ and $B$ are the points of contact of the tangents drawn from the point $P(-3, 1)$ to the circle $x^2+y^2-4x+2y-4=0$,then the equation of the circumcircle of the triangle $PAB$ is

Similar Questions

Given the circles $x^2 + y^2 - 4x - 5 = 0$ and $x^2 + y^2 + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha, \beta)$ such that the lengths of the tangents from $P$ to both circles are equal. Then:

The circle $S \equiv x^2+y^2-2x-4y+1=0$ cuts the $y$-axis at $A, B$ $(OA > OB)$. If the radical axis of $S=0$ and $S^{\prime} \equiv x^2+y^2-4x-2y+4=0$ cuts the $y$-axis at $C$,then the ratio in which $C$ divides $AB$ is:

Two circles centered at $(2,3)$ and $(4,5)$ intersect each other. If their radii are equal,then the equation of the common chord is

$A$ point that lies on the common tangent to the circles $x^2+y^2-2x+18y+78=0$ and $x^2+y^2+8x-6y-200=0$ among the following options is

The length of the common chord of the circles $(x - a)^{2} + y^{2} = c^{2}$ and $x^{2} + (y - b)^{2} = c^{2}$ is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module