Question

How do you write the equation for a circle with center at (4,1) and a radius of 3?

Answer 1

`x^2+y^2-8x-2y+8=0`

Explanation:

Suppose that the centre of a circle is `(x_0,y_0)` and, its radius `r`.

Then, eqn. of the circle is `(x-x_0)^2+(y-y_0)^2=r^2`

Therefore, the reqd. eqn. is, `(x-4)^2+(y-1)^2=3^2`, i.e.,

`x^2+y^2-8x-2y+8=0`