Find the equation of a circle with a center at (7, 2) and a point on the circle at (-2, -5).

Answer:[tex](x-7)^2+(y-2)^2=34[/tex]Step-by-step explanation:We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5). First, recall that the equation of a circle is given by: [tex](x-h)^2+(y-k)^2=r^2[/tex]Where (h, k) is the center and r is the radius. Since our center is at (7, 2), h = 7 and k = 2. Substitute: [tex](x-7)^2+(y-2)^2=r^2[/tex]Next, the since a point on the circle is (2, 5), y = 5 when x = 2. Substitute: [tex](2-7)^2+(5-2)^2=r^2[/tex]Solve for r: [tex](-5)^2+(3)^2=r^2[/tex]Simplify. Thus: [tex]25+9=r^2[/tex]Finally, add: [tex]r^2=34[/tex]We don't need to take the square root of both sides, as we will have the square it again anyways. Therefore, our equation is: [tex](x-7)^2+(y-2)^2=34[/tex]