Question

Find an equation of the circle that has center (6.-3) and passes through (1, 1).

Answer 1

`(x-6)^(2) + (y+3)^(2) =41`

Explanation:

Equation of circle is

`(x-a)^(2) + (y-b)^(2) =r^(2)`

Where (a, b) is center of circle and r is radius

Here (a, b) = (6, -3) and radius can be determined by calculating the distance from center to point (1, 1)

`r=√((6-1)^2 +(-3-1)^2) \\r= √(25+16)\\r= √(41)`

Substituting values in equation of circle

`(x-6)^(2) + (y+3)^(2) =41`