Q. 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) $x^{2} + y^{2} - 6x + 4y + 13 = 0$
Explanation
(c)
Radius of circle=Perpendicular distance from $(3,\ - 2)$ to the line
$4x + 3y + 19 = 0$
$= \frac{4(3) + 3( - 2) + 19}{\sqrt{16 + 9}} = 5$
$\therefore$ Required equation of circle is
${(x - 3)}^{2} + {(y + 2)}^{2} = 5^{2}$
$\Rightarrow \ x^{2} + y^{2} - 6x + 4y - 12 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.