[tex] \text{we know that the general equation of the circle is }\\ \\ (x-h)^2+(y-k)^2=r^2\\ \\ \text{where, (h,k) is the center of the circle and r is the radius.}\\ \\ \text{Given that the centered of the circle is at }(h,k)=(3,5)\\ \text{and a solution point on circle is (-2,17).}\\ \\ \text{so the radius of the circle will be the distance between center and the } [/tex]
[tex] \text{solution point. so using the distance formula, we have}\\ \\ \text{Radius, }r=\sqrt{(-2-3)^2+(17-5)^2}\\ \\ \Rightarrow r=\sqrt{25+144}\\ \\ \Rightarrow r=\sqrt{169}\\ \\ \Rightarrow r=13\\ \\ \text{Hence using the standard for of circle, equation of circle with center}\\ (h,k)=(3,5) \text{ and radius r=13 is}\\ \\ (x-3)^2+(y-5)^2=(13)^2\\ \\ \Rightarrow (x-3)^2+(y-5)^2=169 [/tex]
