Question

How do you find the equation of the circle given center (2, 7) and its radius is 36?

Answer 1

`x^2+y^2-4x-14y-1243=0.`

Explanation:

The eqn. of a circle with centre at pt. `(h,k)` & radius `=r` is
`(x-h)^2+(y-k)^2=r^2.`

Hence, the reqd. eqn. is `(x-2)^2+(y-7)^2=36^2.`

Upon simplification, it becomes `x^2+y^2-4x-14y-1243=0.`