Question

The equation of the circle whose radius is $\sqrt{13}$ and which touches the line $2x-3y+1=0$ at (1,1) is

• A. $(x-3{)}^2+(y-2{)}^2=13$
• B. $(x-3{)}^2+(y+2{)}^2=13$
• C. $(x+4{)}^2+(y+2{)}^2=13$
• D. $(x-1{)}^2+(y+2{)}^2=13$

Answer 1

The correct option is B $(x-3{)}^2+(y+2{)}^2=13$

Let centre of circle is (h,k)

$\sqrt{13}=\left|\frac{2h-3k+1}{\sqrt{13}}\right|$

$13=(2h-3k+1)----\left(1\right)$

$13=2h-3k+1$

$e{q}^n$ of line passing through centre is also passes through (1,1) and is perpendicular to $2x-3y+1=0$

$(y-1)=\frac{-3}{2}(x-1)$

$2y-2x=-3x+3$

$2y+3x=5$

$\therefore 2k+3h=5----\left(2\right)$

From (1),(2) $k=-2$ and $h=3$

$\therefore e{q}^n$ of circle is $(x-3{)}^2+(h+2{)}^2=13$