Question

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

Answer 1

@RadEn

Answer 2

@tcarroll010

Answer 3

so if the center is (2,-5) and one point on the circle is (-2,10), you have to use the distance formula to find the radius of the circle. so:
\[r = \sqrt{(x2-x1)^{2}+(y2-y1)^{2}}\]
putting in numbers: \[r=\sqrt{((2)-(-2))^{2}+((-5)-(10))^{2}}\]
so r is about equal to 15.524

the equation of a circle is "(x-a)^2+(y-b)^2 = r^2" where (a,b) is the center and the radius is "r". can you solve it now?

Answer 4

im sorry im confused =[ can you dumb that down a little more please lol =/ imma slow learner

Answer 5

Would you like me to take you through the whole process, slowly?

Answer 6

yess

Answer 7

please

Answer 8

Okay. First thing we need to do is to find the radius of the circle. That's the distance from the center to the edge.

Let's pretend we have a circle that is centered at the origin. A circle by definition is the set of all points which are the same distance from the center. I've marked a point (x,y) on the drawing and drawn a radius from the center to that point, as you can see.

Answer 9

Now, we know that the radius stays the same, but the triangle that is formed by the radius and the x and y axes is always changing — the values of x and y change as we go around the circle. With me so far?

Answer 10

yes

Answer 11

Okay. That triangle is a right triangle. We know the three sides are x, y, and r. Does the diagram (the triangle, specifically) remind you of anything?

Answer 12

How about the Pythagorean theorem?

Answer 13

The hypotenuse is \(r\). The length of the base is \(x\), and the height is \(y\). That means if we have a circle of radius \(r\) with its center at (0,0), we can find any point on the circle with the equation \(x^2+y^2=r^2\).

Answer 14

Now what if we didn't have the center at (0,0)? Well, for one thing, it would be slightly trickier to find out the length of each side of the triangle. Not too tricky, though. We had the simple case with one end of our line being at (0,0), so if we went along the x-axis to x, we knew that we were x units away from 0. What if we started at 2? Well, then we would be x-2 units away. Same story for y. So, we can generalize a bit with a new drawing: