Question

What is the equation of the circle with center (0, 1) and a radius of 10?

Answer 1

Answer:  

The equation of the circle with center (0, 1) and a radius of 10 is $\bold{x^{2} + y^{2} - 2y-99 = 0}$

Solution:

Equation of circle whose center (h, k) and radius r is given as,

$(x-h)^{2} + (y-k)^{2} = r^{2}$ ---- eqn 1

From question, given that the circle has radius 10 and center (0, 1). We have to find the equation of the circle.

Hence we get, h = 0 ; k = 1; r = 10

By using eqn 1, the equation of circle whose center (0, 1) and radius 10 is given as

$(x-0)^{2} + (y - 1)^{2} = 10^{2}$

On Simplifying we get

$x^{2} + (y-1)^{2} = 10^{2}$

Expand $(y-1)^{2}$ using the formula $(a-b)^{2} = a^{2} - 2ab + b^{2}$

$x^{2} + y^{2} - 2y + 1 = 10^{2}$

$x^{2} + y^{2} - 2y + 1 = 100$

$x^{2} + y^{2} - 2y + 1-100 = 0$

$x^{2} + y^{2}- 2y- 99 = 0$

Hence equation of the circle with center (0, 1) and a radius of 10 is  $\bold{x^{2}+y^{2}-2 y-99=0}$