Question

Fine the equation of a circle that has a diameter with end points $(-6,1)$ and $(2,-5)$

• A. ${(x+2)}^2+{(y+2)}^2=25$
• B. $(x-2{)}^2+(y-2{)}^2=25$
• C. $(x+5{)}^2+(y+1{)}^2=36$
• D. $(x+4{)}^2+(y+4{)}^2$

Answer 1

The correct option is A ${(x+2)}^2+{(y+2)}^2=25$

We need to find both the center and the radius.

To find the center $C$ with end points $(-6,1)$ and $(2,-5)$, we will use the midpoint formula since the center must lie equidistant from the two given points.

$C=(\frac{x_1{+x}_2}{2},\frac{y_1+y_2}{2})=(\frac{-6+2}{2},\frac{1-5}{2})=(\frac{-4}{2},\frac{-4}{2})=(-2,-2)$

Next, to find the radius we will use the distance formula on the center $(h,k)=(-2,-2)$ and either one of the given points. We wil use $(2,-5)$ as shown below:

$r=\sqrt{{\left(x_1{-x}_2\right)}^2+{\left(y_1{-y}_2\right)}^2}$

$=\sqrt{{(-2-2)}^2+{(-2+5)}^2}$

$=\sqrt{{(-4)}^2+{\left(3\right)}^2}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$

Finally, since we know that $(h,k)=(-2,-2)$ and $r=5$, substituting into the equation of a circle we find that

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

$\Rightarrow {(x-(-2\left)\right)}^2+{(y-(-2\left)\right)}^2={\left(5\right)}^2$

$\Rightarrow {(x+2)}^2+{(y+2)}^2=25$

Hence, the equation of the circle is ${(x+2)}^2+{(y+2)}^2=25$.