SOLUTION:
Step 1:
In this question, we are asked to find the equation of the circle with the given characteristics in standard form:
Step 2:
We need to get the distance between the two points:
(11 , - 7 ) and ( 17 , 13 )
[tex]\begin{gathered} d\text{ =}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ where\text{ \lparen x}_1,\text{ y}_1)\text{ = \lparen 11, - 7\rparen} \\ \text{and } \\ (x_2,\text{ y}_2)\text{ = \lparen 17, 13\rparen} \end{gathered}[/tex][tex]\begin{gathered} d=\sqrt{(17-11)^2+\text{ \lparen13-\lparen-7\rparen\rparen}^2} \\ d\text{ = }\sqrt{6^2+\text{ \lparen13+7\rparen}^2} \\ d\text{ =}\sqrt{36+400} \\ d\text{ =}\sqrt{436} \\ Radius\text{ =}\frac{\sqrt{436}}{2} \end{gathered}[/tex]
Next:
[tex]\begin{gathered} Using\text{ the equation of a circle in standard form, we have that:} \\ (x-a)^2+\text{ \lparen y-b\rparen}^2\text{ = r}^2 \\ where\text{ \lparen a , b \rparen= \lparen}\frac{11+17}{2},\text{ }\frac{-7+13}{2})\text{ = \lparen}\frac{28}{2},\text{ }\frac{6}{2})\text{ = \lparen 14, 3\rparen} \\ r=\frac{\sqrt{436}}{2} \\ and \\ r^2\text{ =\lparen}^\frac{\sqrt{436}}{2})^2=\frac{436}{4}=\text{ 109} \end{gathered}[/tex][tex]\begin{gathered} (x-14)^2+\text{ \lparen y - 3\rparen}^2=\text{ 109} \\ (Equation\text{ of the circle in standard form\rparen} \\ \end{gathered}[/tex]
The graph is as follows:
