Question

The equation of a circle with center (5,-3) and radius of 9

Answer 1

`\large\boxed{x^2+y^2-10x+6y-47=0}`

Explanation:

The equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

(h, k) - center

r - radius

We have the center (5, -3) and the radius r = 9.

Substitute:

`(x-5)^2+(y-(-3))^2=9^2\\\\(x-5)^2+(y+3)^2=81`

Use

`(a\pm b)^2=a^2\pm2ab+b^2`

`x^2-2(x)(5)+5^2+y^2+2(y)(3)+3^2=81\\\\x^2-10x+25+y^2+6y+9=81\\\\x^2+y^2-10x+6y+(25+9)=81\\\\x^2+y^2-10x+6y+34=81\qquad\text{subtract 81 from both sides}\\\\x^2+y^2-10x+6y-47=0`