Q. The equation of the circle with centre (2, 1) and touching the line $3x + 4y = 5$ is
A) $x^{2} + y^{2} - 4x - 2y + 5 = 0$
B) $x^{2} + y^{2} - 4x - 2y - 5 = 0$
C) $x^{2} + y^{2} - 4x - 2y + 4 = 0$
D) $x^{2} + y^{2} - 4x - 2y - 4 = 0$
Explanation
(c)
Distance from centre (2, 1) to the line $3x + 4y - 5 =$ radius of circle
$\Rightarrow \ \frac{|3(2) + 4(1) - 5|}{\sqrt{3^{2} + 4^{2}}} = r\ \ \ \Rightarrow \ \ r = 1$
$\therefore$ Equation of circle is
$(x - 2)^{2} + (y - 1)^{2} = 1^{2}$
$\Rightarrow \ x^{2} + y^{2} - 4x - 2y + 4 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.