The line x+y+2=0 intersects the circle x^2+y^ | vedclass

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

Question

(B) The equation of any circle passing through the intersection of the line $x+y+2=0$ and the circle $x^2+y^2+4x-4y-4=0$ is given by $x^2+y^2+4x-4y-4+

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The equation of any circle passing through the intersection of the line $x+y+2=0$ and the circle $x^2+y^2+4x-4y-4=0$ is given by $x^2+y^2+4x-4y-4+\lambda(x+y+2)=0$. Rearranging the terms,we get $x^2+y^2+(4+\lambda)x+(\lambda-4)y+(2\lambda-4)=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we have $2g = 4+\lambda$,$2f = \lambda-4$,and $c = 2\lambda-4$. The centre of the circle $S$ is $(-g, -f) = \left(-\frac{4+\lambda}{2}, -\frac{\lambda-4}{2}\right)$. The distance of the centre $(-g, -f)$ from the line $x+y+2=0$ is given as $\sqrt{2}$. Using the distance formula $\left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right| = d$,we get $\left|\frac{-g-f+2}{\sqrt{1^2+1^2}}\right| = \sqrt{2}$. This implies $|-g-f+2| = 2$,so $-g-f+2 = 2$ or $-g-f+2 = -2$. Case $1$: $g+f = 0$. Substituting $g$ and $f$,$\frac{4+\lambda}{2} + \frac{\lambda-4}{2} = 0 \Rightarrow \lambda = 0$. If $\lambda=0$,the circle is the original circle $x^2+y^2+4x-4y-4=0$,but the problem states $S$ is a different circle. Case $2$: $g+f = 4$. Substituting $g$ and $f$,$\frac{4+\lambda}{2} + \frac{\lambda-4}{2} = 4 \Rightarrow \lambda = 4$. For $\lambda=4$,$g = \frac{4+4}{2} = 4$,$f = \frac{4-4}{2} = 0$,and $c = 2(4)-4 = 4$. Thus,$g+f+c = 4+0+4 = 8$.

The line $x+y+2=0$ intersects the circle $x^2+y^2+4x-4y-4=0$ at two points $A$ and $B$. Let $S \equiv x^2+y^2+2gx+2fy+c=0$ be a different circle passing through the points $A$ and $B$. If the distance of the centre of $S=0$ from $AB$ is $\sqrt{2}$,then $g+f+c=$

Similar Questions

If $x-4=0$ is the radical axis of two orthogonal circles,out of which one is $x^2+y^2=36$,then the centre of the other circle is

If the circles $x^{2}+y^{2}+6x+8y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y = k+6\sqrt{3}+8\sqrt{6}$ with $k>0$ touch internally at the point $P(\alpha, \beta)$,then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$

Consider the family of circles $x^2+y^2-2x-2\lambda y-8=0$. This family passes through two fixed points $A$ and $B$. Find the distance between these two points.

If the circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ touch each other internally,then $\alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module