Question

If $ (3,-2) $ is the centre of a circle and $ 4x+3y+19=0 $ is a tangent to the circle, then the equation of the circle is

• A. $ {{x}^{2}}+{{y}^{2}}-6x+4y+25=0 $
• B. $ {{x}^{2}}+{{y}^{2}}-6x+4y+12=0 $
• C. $ {{x}^{2}}+{{y}^{2}}-6x+4y-12=0 $
• D. $ {{x}^{2}}+{{y}^{2}}-6x+4y+13=0 $
• E. $ {{x}^{2}}+{{y}^{2}}-6x+4y+9=0 $

Answer 1

Answer: (C)

Radius of circle = Perpendicular distance from
$ (3,-2) $ to the line $ 4x+3y+19=0 $ and
$=\frac{4(3)+3(-2)+19}{\sqrt{16+9}}=5 $
$ \therefore $ Required equation of circle is $ {{(x-3)}^{2}}+{{(y+2)}^{2}}={{5}^{2}} $
$ \Rightarrow $ $ {{x}^{2}}+{{y}^{2}}-6x+4y-12=0 $