Question

What is the equation of a circle with center of (4,2) passing through the point (7,2)

Answer 1

Answer:

(x-4)^2 + (y-2)^2 = 3^2

or

(x-4)^2 + (y-2)^2 = 9 (simplified if needed)

Step-by-step explanation:

-Equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$ where the center is (h, k) and the radius is $r^2$.

-Place the center and the point onto that equation:

$(7-4)^2+(2-2)^2=r^2$

-Then, you solve:

$(7-4)^2+(2-2)^2=r^2$

$(3)^2+(0)^2=r^2$

$9 +0=r^2$

$9=r^2$

$\sqrt{9} =\sqrt{r^2}$

$3=r$

-So, the result is:

$(x-4)^2+(y-2)^2=3^2$

or

$(x-4)^2+(y-2)^2=9$ (simplified if needed)