Question

A circle passes through point $(3,\sqrt{\frac{7}{2}})$ touches the line pair $x^2-y^2-2x+1=0$ . Centre of circle liesinside the circle $x^2+y^2-8x+10y+15=0.$ Co-ordinate of centre of circle is

• A. $(4,0)$
• B. $(5,0)$
• C. $(6,0)$
• D. $(0,4)$

Answer 1

Answer: (A)

$(4,0)$

$\Rightarrow$$(x-1+y)\cdot (x-1-y)$=$0$

$\Rightarrow x+y-1$=$0$ and $x-y-1$=$0$

Circle touches above two lines which are symmetrical about $x$ -axis.

-. Centre of the circle lies on $x$ -axis

$\Rightarrow \left|\frac{h-0-1}{\sqrt{2}}\right|=\sqrt{(h-3{)}^2+7/2}$

$\Rightarrow h^2+1-2h=2[n^2+9-6h+\frac{7}{2}]$

$\Rightarrow h^2-10h+24=0$

$\Rightarrow h=6,4$

Hence required point is (4,0)