Question

Find the equation of a family of circles touching the lines $x^2-y^2+2y-1=0$. (where $h$ and $k$ are parameters)

• A. $x^2+y^2-2ky+\frac{k^2}{2}+k-\frac{1}{2}=0$
• B. $x^2+y^2-2hx-2y+\frac{h^2}{2}+1=0$
• C. $x^2+y^2-2hx+2y+\frac{h^2}{2}=0$
• D. $x^2+y^2-2ky+\frac{k^2}{2}=0$

Answer 1

Answer: (A), (B)

The correct options are
A $x^2+y^2-2ky+\frac{k^2}{2}+k-\frac{1}{2}=0$
B $x^2+y^2-2hx-2y+\frac{h^2}{2}+1=0$

Given pair of lines is
$x^2-y^2+2y-1=0$
$(x-y+1)(x+y-1)=0$

The centre of family of circles touching the above lines will lie on the angle bisectors of the above lines.

Equations of the angle bisectors of the above lines are given by
$x=0$ and $y=1$

Case $1:$
Let $A(h,1)$ be a point on the line $y=1,AP=\frac{h}{\sqrt{2}}$
Equation of circle touching the given line is $(x-h{)}^2+(y-1{)}^2=\frac{h^2}{2}$
i.e., $x^2+y^2-2hx-2y+\frac{h^2}{2}+1=0$

Case $2:$
Let $B(0,k)$ be a point on the line $x=0$
Now, $BN=\frac{k-1}{\sqrt{2}}$
Equation of circle touching the given lines is
$(x-0{)}^2+(y-k{)}^2=\frac{(k-1{)}^2}{2}$
or $x^2+y^2-2ky+\frac{k^2}{2}+k-\frac{1}{2}=0$