Question

what is the equation of a circle with the center at (-3,1) and through the point (2, 13)

Answer 1

Remember the standard equation of a circle from the previous problem?

Answer 2

\( (x - h)^2 + (y - k)^2 = r^2 \)

Answer 3

ok so h= -3 and k=1

Answer 4

Right.
Let's put that in the equation:

\((x + 3)^2 + (y - 1)^2 = r^2\)

Ok?

Answer 5

Now take the given point, (2, 13), and replace x with 2 and y with 13. Then you can find r.
Then rewrite the equation using the r value you find.

Answer 6

so (2+3)^2 + (13-1)^2 =11^2 right?

Answer 7

How do you know r = 11? That's what you are solving for.

Answer 8

(2+3)^2 + (13-1)^2= r^2

Answer 9

\( (2+3)^2 + (13-1)^2 =r^2\)

Solve for r.

Answer 10

Correct.
Calculate the number on the left side and find r.

Answer 11

ok so (2+3)^2=11 or do i do it 2+3 they square it making it 25

Answer 12

Do what is in parentheses first. Then square.

Answer 13

so it would be r= 25^2

Answer 14

What happened to (13 - 1)^2?

Answer 15

\( (2+3)^2 + (13-1)^2 =r^2\)

\(5^2 + 12^2 = r^2 \)

\(25 + 144 = r^2\)

\(169 = r^2\)

r = ?

Answer 16

13

Answer 17

Good. Now that we know r, we can write the final solution:

The equation of the circle is

\((x+3)^2 + (y-1)^2 =169\)