Chapter 15, Problem 30CR

To determine

To Find: Theequation of circle.

Answer to Problem 30CR

  ${\left(x-2\right)}^2+{\left(y-4\right)}^2=10$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Center of the circle}=\left(2,4\right)\\ \text{And Passes through point }\left(1,7\right)\end{array}$

Formula Used:

  $\text{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation:

  $\begin{array}{l}\left(h,k\right)=\left(2,4\right)\\ \text{And Passes through point }\left(1,7\right)\\ \text{The distance between the center of the circle and the given point}\\ \text{will be the radius of the circle so,}\\ radius=\sqrt{{\left(1-2\right)}^2+{\left(7-4\right)}^2}=\sqrt{1+9}=\sqrt{10}\\ {\left(x-2\right)}^2+{\left(y-4\right)}^2={\left(\sqrt{10}\right)}^2\\ {\left(x-2\right)}^2+{\left(y-4\right)}^2=10\end{array}$

Conclusion:

Thus, the required equation of circle is ${\left(x-2\right)}^2+{\left(y-4\right)}^2=10$ .