Question

What is the equation of the circle with center (2,-5) that passes through the point (-2,10)?

a. (x-2)^2+(y-(-5))^2=25
b. (x-(-2))^2+(y-10)^2=241
c. (x-2)^2+(y-(-5))^2=241
d. (x-(-2))^2+(y-10)^2=25

I'm not just asking for an answer. I don't understand and any form of help would be appreciated :)

Answer 1

Do you know about center part?

Answer 2

Saying we have center \((h,k)\), then equation would be \((x-h)^2+(y-k)^2 = r^2\)

Right?

Answer 3

I think that I might understand what you're getting at.. are you implying that the answer would be "a. (x-2)^2+(y-(-5))^2=25" or would it be "c. (x-2)^2+(y-(-5))^2=241" ...?

Answer 4

Yeah, we are either at A or C, we don't know yet because we haven't figure out the radius part.

Answer 5

\(r\) in \((x-h)^2+(y-k)^2 = r^2\) stands for radius.

You can find radius by using distance formula on points \((2,-5)\) and \((-2,10)\)

Answer 6

Distance formula is \[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Answer 7

You can use directly the general equation of a circle to get r by replace the point into the formula