Question

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

Answer 1

Answer:

$(x-6)^{2} + (y+3)^{2} =41$

Step-by-step 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)  

$r=\sqrt{(6-1)^2 +(-3-1)^2} \\r= \sqrt{25+16}\\r= \sqrt{41}$

Substituting values in equation of circle

$(x-6)^{2} + (y+3)^{2} =41$