Question

A variable circle always touches the line $y=x$ and passess-through the point $(0,0)$. The common chords of above circle and $x^2+y^2+6x+8y-7=0$ will pass through a fixed point whose co-ordinates are

• A. $(1,1)$
• B. $(2,2)$
• C. $(1/2,1/2)$
• D. $noneofthese$

Answer 1

The correct option is C $(1/2,1/2)$

The family of circle touching a line lx + my + n = 0 at a given point (h,k) on it is

$(x-h{)}^2+(y-k{)}^2+\lambda (lx+my+n)=0$ where $\lambda \in R$

So, for given conditions equation of circle will be of form,

$(x-0{)}^2+(y-0{)}^2+\lambda (x-y)=0$

$x^2+y^2+\lambda x-\lambda y=0$

We know that common chord of circles with equation $S_1$ and $S_2$ is given by $S_1-S_2=0$

Now, let's find the common chord of two circles,

$\to (x^2+y^2+\lambda x-\lambda y)-(x^2+y^2+6x+8y-7)=0$

$\to (\lambda -6)x-(\lambda +8)y+7=0$

It is clearly evident that above equation is always valid for point $(\frac{1}{2},\frac{1}{2}).$

Hence, (C) is the correct answer.