# Equation of a circle with specific conditions

I am trying to find the equation for a circle that has the center at (-1 , 4) and passed through the point (3, -2)

This seems like a straightforward problem.

$$(x+1)^2+(y-4)^2 = 16$$

The radius should just be the difference between the x points squared but that is wrong. Why?

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The radius should be the distance between (-1,4) and (3,-2), but it's not what computed. –  ՃՃՃ Dec 3 '12 at 1:46
@use I figured that out, the distance formula gives $\sqrt{8}$ which is still wrong. –  user138246 Dec 3 '12 at 1:50
The distance is $\sqrt{52}$, which gives you the equation $(x+1)^2+(y-4)^2=52$. –  ՃՃՃ Dec 3 '12 at 1:55

To find the equation for a circle that has the center at $(-1 , 4)$ and passes through the point $(3, -2)$, the right hand side of your equation is fine: you used the center of the circle correctly. So we have: $$(x+1)^2+(y-4)^2 = r^2$$

To determine the radius of the circle, you need to compute the distance between the given point on the circumference of the circle and the center of the circle:

$$r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(3 - (-1))^2 + (-2 - 4)^2} = \sqrt{4^2+ (-6)^2} =\sqrt{52}$$

Since we need to square $r$, we can use $r^2 = 52$.

$$(x+1)^2+(y-4)^2 = 52$$

(I simply chose $(3, -2) = (x_1, y_1)$ and $(-1, 4) = (x_2, y_2)$, but you get the same result if the points were assigned differently. Since the difference between the $x$-coordinates and the difference between the $y$-coordinates are both squared, which points you take as point 1 and point 2 is of no concern.)

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