Question

Write the equation of the circle with center (-2, -5) and containing the point (-10, -20).

I know the first part will be (x+2)^2+(y+5)^2=? I don't know the radius.

Answer 1

Well, you're correct on the first part, and the added fact is that they give you a point on the circle (an x,y to plug in)

Answer 2

\[\Large \text{The equation of a circle}=\]
\[\Large (x-h)^2+(y-k)^2=r^2\]

Answer 3

They give us h,k which is -2,-5 and then they give us the x and y which is -10,-20 so we plug that in and then take the sqrt to get the r

Answer 4

plug what in where??

Answer 5

You told me you have (x+2)^2+(y+5)^2=?
right?
well the x part of the equation refers to any x-value on this circle and the y in that equation refers to any y-value of this circle. We're fortunate enough that they give us a point (which is an x/ y value on our circle)

The point is -10, -20 and so we plug that in for the x and y for the equation of a circle that you put before

(x+2)^2+(y+5)^2=?

Answer 6

great start.
if you sketch the points you will see more.
use the distance formula (pythagorean theorem) to get the radius

Answer 7

So I would have (-10+2)^2+(-20+5)^2=289

Answer 8

so does that mean my equation would be (x+2)^2+(y+5)^2=289??

Answer 9

Yes! And in our eqaution our result gives us r^2 which is the radius squared

Answer 10

i did not actually do the problem. sq rt of 289 = 17

Answer 11

but you have to square the radius to get the right answer @triciaal

Answer 12

Yes, good job! @MLeighW

Answer 13

you are correct! was just thinking in terms of the formula.