The circle S=0 cuts the circle x^2+y^2-4x+2y- | vedclass

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

Question

(A) Given that,the circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally and the centre of the circle $S=0$ is $(2,3)$.Let the equation of the

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given that,the circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally and the centre of the circle $S=0$ is $(2,3)$.Let the equation of the circle $S=0$ be $x^2+y^2+2gx+2fy+c=0$. Since the centre is $(-g, -f) = (2,3)$,we have $g=-2$ and $f=-3$.As we know that,if two circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g'x+2f'y+c'=0$ intersect orthogonally,then $2gg'+2ff'=c+c'$.Here,$(g, f) = (-2, -3)$ and for the given circle $x^2+y^2-4x+2y-7=0$,$(g', f') = (-2, 1)$ and $c' = -7$.Substituting these values into the condition:$2(-2)(-2) + 2(-3)(1) = c + (-7)$$8 - 6 = c - 7$$2 = c - 7$$c = 9$The radius $r$ of the circle $S=0$ is given by $\sqrt{g^2+f^2-c}$.$r = \sqrt{(-2)^2 + (-3)^2 - 9} = \sqrt{4 + 9 - 9} = \sqrt{4} = 2$.

The circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally. If $(2,3)$ is the centre of the circle $S=0$,then its radius is

Similar Questions

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally, then $k$ is equal to

The equation of a circle which touches the straight lines $x+y=2$,$x-y=2$ and also touches the circle $x^2+y^2=1$ is

The distance of the origin from the external centre of similitude for the circles $x^2+y^2-8x-10y-8=0$ and $x^2+y^2+2x-2y-2=0$ is

Two circles $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut each other orthogonally,then:

If the centre of a circle,which passes through the points of intersection of the circles $x^2 + y^2 - 6x + 2y + 4 = 0$ and $x^2 + y^2 + 2x - 4y - 6 = 0$,lies on the line $y = x$,then the equation of the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module