A circle S passes through the points of inter | vedclass

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

Question

(D) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + kS_2 = 0$.$(x^2+y^2-2

Vedclass · Accepted Answer

$\sqrt{14}$

Vedclass · Answer

(D) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + kS_2 = 0$.$(x^2+y^2-2x+2y-2) + k(x^2+y^2+2x-2y+1) = 0$$(1+k)x^2 + (1+k)y^2 + (2k-2)x + (2-2k)y + (k-2) = 0$Dividing by $(1+k)$,we get $x^2 + y^2 + \frac{2(k-1)}{k+1}x + \frac{2(1-k)}{k+1}y + \frac{k-2}{k+1} = 0$.The centre of this circle is $\left(-\frac{k-1}{k+1}, -\frac{1-k}{k+1}\right) = \left(\frac{1-k}{k+1}, \frac{k-1}{k+1}\right)$.Since the centre lies on the line $x-y+6=0$,we substitute the coordinates:$\frac{1-k}{k+1} - \frac{k-1}{k+1} + 6 = 0$$\frac{1-k-k+1}{k+1} = -6 \Rightarrow 2-2k = -6k-6$$4k = -8 \Rightarrow k = -2$.Substituting $k=-2$ into the equation: $x^2+y^2 + \frac{2(-3)}{-1}x + \frac{2(3)}{-1}y + \frac{-4}{-1} = 0 \Rightarrow x^2+y^2+6x-6y+4=0$.The radius $r = \sqrt{g^2+f^2-c} = \sqrt{3^2+(-3)^2-4} = \sqrt{9+9-4} = \sqrt{14}$.

$A$ circle $S$ passes through the points of intersection of the circles $x^2+y^2-2x+2y-2=0$ and $x^2+y^2+2x-2y+1=0$. If the centre of this circle $S$ lies on the line $x-y+6=0$,then the radius of the circle $S$ is

Similar Questions

If the circles $x^2 + y^2 - 2ax + c = 0$ and $x^2 + y^2 + 2by + 2\lambda = 0$ intersect orthogonally,then the value of $\lambda$ is

The number of common tangents to the circles $x^2+y^2+4x=0$ and $x^2+y^2-2x=0$ is

If a circle has its centre on the line $x-y-1=0$ and passes through the points of intersection of the two circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y-7=0$,then the centre of that circle is

Let the set of all values of $r$, for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points, be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

The equation of the radical axis of the circles $2x^2 + 2y^2 - 7x = 0$ and $x^2 + y^2 - 4y - 7 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module