Chapter 9.2, Problem 37WE

To determine

To find: The equation of the circle.

Answer to Problem 37WE

The required equation is ${\left(x-1\right)}^2+{\left(y-4\right)}^2=2$ .

Explanation of Solution

Given information: It is given that the coordinates of a diameter are $\left(2,5\right)and\left(0,3\right)$ .

Formula used: $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2},M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ and ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: As we know that the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ has the center $\left(h,k\right)$ and radius $\left(r\right)$ .

Firstly, calculate the mid-point of the diameter. i.e. center.

  $\begin{array}{l}Center\left(O\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ O=\left(\frac{2+0}{2},\frac{5+3}{2}\right)\\ O=\left(1,4\right)\end{array}$

Now, get the radius by applying the distance formula. The 2 points are $\left(1,4\right)and\left(2,5\right)$ .

  $\begin{array}{l}r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(2-1\right)}^2+{\left(5-4\right)}^2}\\ =\sqrt{1+1}\\ \therefore r=\sqrt{2}\end{array}$

Now, the equation of the circle can be written,

  $\begin{array}{l}\therefore {\left(x-1\right)}^2+{\left(y-4\right)}^2=2\left\{\because \left(h,k\right)=\left(1,4\right);r=\sqrt{2}\right\}\\ \therefore {\left(x-1\right)}^2+{\left(y-4\right)}^2=2\end{array}$

Therefore, the required equation is ${\left(x-1\right)}^2+{\left(y-4\right)}^2=2$ .

The graph is given below,