Question

A circle passing through the point {2, 2($\sqrt{2}$ – 1)} touches the pair of lines x² – y² – 4x + 4 = 0. The centre of the circle is –

• A. (2, 2$\sqrt{2}$) & (2, 6$\sqrt{2}$ – 8)
• B. (2, 5$\sqrt{2}$) & (2, 7$\sqrt{2}$)
• C. (2, 5$\sqrt{2}$ – 1) & (2, –3)
• D. None of these

Answer 1

Answer: (A)

(2, 2$\sqrt{2}$) & (2, 6$\sqrt{2}$ – 8)

Answer 2

The equation of the pair of lines is

(x – 2)² – y² = 0

i.e. (x – 2) = ± y

i.e. x – y = 2 and x + y = 2

The centre of the circle touching the above lines must lie on the angular bisectors of the above lines. Hence, C ŗ (2, k)

(see figure)

Thus we have CM = CP

i.e. $\frac{|2 + k - 2|}{\sqrt{2}}$ = 2($\sqrt{2}$– 1) – k

i.e. ± $\frac{k}{\sqrt{2}}$ = 2($\sqrt{2}$– 1) – k

i.e. k $\left( 1 \pm \frac{1}{\sqrt{2}} \right)$ = 2($\sqrt{2}$– 1)

gives k = $\frac{2\sqrt{2}(\sqrt{2} - 1)}{\sqrt{2} \pm 1}$ =$2\sqrt{2}$, $2\sqrt{2}$ ($\sqrt{2}$– 1)²

=$2\sqrt{2}$, $6\sqrt{2}$– 8.