Chapter 1, Problem 50RE

To determine

To find:The center, radius and the standard equation of the circle.

Answer to Problem 50RE

The center of the circle is (1,-2), the radius is $\sqrt{32}$ and the standard form of the equation of the circle is ${\left(x-2\right)}^2+{\left(y+2\right)}^2=32$ .

Explanation of Solution

Given: $\left(-3,2\right)$ and $\left(5,-6\right)$ .

Formula Used:

TheMidpoint formula is $M=\left(x,y\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

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

Calculation:

The center of a circle $\left(h,k\right)$ is equidistant from any point on the circle. The center of the given circle is the midpoint of the line joining the points $\left(-3,2\right)$ and $\left(5,6\right)$ .

Use the midpoint formula to find the other point.

  $M=\left(x,y\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Substitute $x_1=-3,y_1=2,x_2=5,y_2=6$ in the formula.

  $\left(x,y\right)=\left(\frac{-3+5}{2},\frac{2+\left(-6\right)}{2}\right)$

  $=\left(\frac{2}{2},\frac{-4}{2}\right)$

  $=\left(1,-2\right)$

Thus, the center of the circle is $\left(1,-2\right)$ .

If $\left(x,y\right)$ represents the coordinates of any point on the circle with radius $r$ and center $\left(h,k\right)$ , then by the distance formula, the radius of the circle is equivalent to the distance between the points $\left(x,y\right)$ and $\left(h,k\right)$ .

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

Consider the point (-3, 2). Substitute $x=-3,h=1,y=2$ and $k=-2$

  $r=\sqrt{{\left(-3-1\right)}^2+{\left[2-\left(-2\right)\right]}^2}$

  $=\sqrt{16+16}$

  $=\sqrt{32}$

The radius is $\sqrt{32}$ .

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

Substitute for $h$ , $k$ , and $r$ .

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

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

Therefore, the standard form of the equation of the circle is ${\left(x-2\right)}^2+{\left(y+2\right)}^2=32$ .

Conclusion:

Therefore, the center of the circle is $\left(1,-2\right)$ , the radius is $\sqrt{32}$ and the standard form of the equation of the circle is ${\left(x-2\right)}^2+{\left(y+2\right)}^2=32$ .