Question

Find the equation of circle which pass through (5, 9) and center at (2, 5).

(a) x^2+y^2+4x-10y+4=0

(b) x^2+y^2-4x-10y+4=0

(c) x^2+y^2+4x+10y+4=0

(d) x^2+y^2+4x-10y-4=0

I have been asked this question by my school principal while I was bunking the class.

Origin of the question is Conic Sections topic in section Conic Sections of Mathematics – Class 11

Answer 1

Right choice is (b) x^2+y^2-4x-10y+4=0

Explanation: Equation of circle with center at (a, b) and radius r units is

(x-a)^2 + (y-b)^2 = r^2

(5-2)^2 + (9-5)^2 = r^2 => r^2=3^2+4^2 => r=5.

So, equation of circle is (x-2)^2+(y-5)^2=5^2 => x^2+y^2-4x-10y+4=0.