Mathematics: What’s the equation of a circle with a center of (-3,12) that passes through the point (3,4)

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What’s the equation of a circle with a center

kbmom1202

26.01.2022

(x+3)^2 +(y-12)^2 = 10^2

(x+3)^2 + (y-12)^2 = 100 (if simplified radius needed).

Step-by-step explanation:

-The equation of a circle is:

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

-Put the center and the point onto that equation, in order to solve and get the equation:

(3+3)^2+(4-12)^2=r^2

-Then, you solve:

(3+3)^2+(4-12)^2=r^2

(6)^2+(-8)^2=r^2

36+64=r^2

100=r^2

$\sqrt{100} =\sqrt{r^2}$

10=r

so, the result can be:

(x+3)^2+(y-12)^2=10^2

(x+3)^2+(y-12)^2=100 (if simplified radius needed).

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