Question

How do you write an equation of a circle with its center in quadrant I, radius of 5 units and tangents at x=2 and y=3?

Answer 1

`x^2+y^2-14x-16y+88=0`

Explanation:

The lines `x=2` and `y=3` intersect at `(2,3)`

and divide Quadrant I, in four parts. As radius is `5`, we can have circle only in right hand top portion

and hence center could be only at `(7,8)` and equation of circle is

`(x-7)^2+(y-8)^2=25`

or `x^2-14x+49+y^2-16y+64=25`

or `x^2+y^2-14x-16y+88=0`
graph{(x-2)(y-3)(x^2+y^2-14x-16y+88)=0 [-15.33, 24.67, -2.68, 17.32]}