Question

Help! This is a Geometry question for a test...

Answer 1

Answer:

a. (-3,2)

b. $\sqrt{65}$

c. $(x+3)^{2} +(y-2)^{2} = 65$

Step-by-step explanation:

To find the center of the circle use the midpoint formula:

$(\frac{x1+x2}{2},\frac{y1+y2}{2} )$

x1 = -10, x2 = 4, y1 = -2, y2 = 6

Plug these into the equation and solve:

$(\frac{-10+4}{2},\frac{-2+6}{2} )$

$(\frac{-6}{2}, \frac{4}{2} )$                           Add the numerators

(-3, 2)                            Divide both by 2

So, the center of the circle is (-3,2)

To find the radius use the distance formula:

$Distance = \sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$

Plug in the values from above and solve:

$r = \sqrt{((-10)-(-3))^{2} +((-2)-2)^{2} }$

$r = \sqrt{(-7)^{2}+(-4)^{2} }$                               Simplify whats in parentheses

$r = \sqrt{49+16}$                                          Distribute the squares

$r=\sqrt{65}$

The radius is $\sqrt{65}$ (this is in radical form).

The equation for a circle is:

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

(h, k) is the center point, which is (-3,2)

r = radius, which is $\sqrt{65}$

Plug everything in, and simplify:

$(x-(-3))^{2}+(y-(2))^{2} = (\sqrt{65} )^{2}$

$(x+3)^{2} +(y-2)^{2} = 65$                               Simplify

Finally, the equation of the circle is $(x+3)^{2} +(y-2)^{2} = 65$.

Hope this helps!! Ask questions if you need!