Chapter 1, Problem 35RE

To determine

To find: The center and radius of the circle. Write the general equation of this circle.

Answer to Problem 35RE

Solution:

Center $\left(1, -2\right)$, radius $4\sqrt{2};x^2 + y^2 - 2x + 4y - 27 = 0$

Explanation of Solution

Given:

The endpoints of the diameter of a circle are $\left(-3, 2\right)\text{and}\left(5, -6\right)$.

The distance between endpoints of the diameter of a circle is $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\sqrt{{\left(5-\left(-3\right)\right)}^2+{\left(-6-2\right)}^2}$

$=\sqrt{64+64}$

$=\sqrt{128}=8\sqrt{2}$

The diameter of the circle is $8\sqrt{2}$

The radius of the circle is $4\sqrt{2}$

To find center of the circle, we find midpoint s between the endpoints of the diameter of a circle are $\left(-3, 2\right)\text{and}\left(5, -6\right)$

Midpoints $=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)=\left(\frac{-3+5}{2}, \frac{2+\left(-6\right)}{2}\right)=\left(1, -2\right)$

Center of the circle is $\left(1, -2\right)$

The standard form of the equation of the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

${\left(x-1\right)}^2+{\left(y-\left(-2\right)\right)}^2={\left(4\sqrt{2}\right)}^2$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=32$

$x^2-2x+1+y^2+4y+4=32$

$x^2+y^2-2x+4y-27=0$