The center of a circle is located at (9, 7). The circle also passes through (-1, -5). What is the equation of this circle in standard form?

Question

[Enter your answer as (x - h)² + (y - k)² = r².]

Answer 1

standard form=(x-h)^2+(y-k)^2=r^2 where the center is (h, k) and the coordinate point is (x,y). Then do this

(x-9)^2+(x-7)^2=r^2 so we need to find r squared and to do that we need to work backwards. first factor out

(x-9)^2 and (x-7)^2. (x-9)^2 becomes (x^2-18x+81) and (x-7)^2 becomes(x^2-14x+49) and when these equations would have completed the square the constants in these factored quadratic expressions would be added together like this (x^2-18x+81)+(x^2-14x+49)=81+49 due to adding these constants on both sides. 81+49 is 130 which equals r^2 so sqrt(130), due to not having a root that's a whole number, is just the radius so the final equation is (x-9)^2+(x-7)^2=sqrt(130)

Answer 2

Use the distance formula to determine the radius. Square and insert in formula.