You have a center of a circle $M(-2,3)$ going through a point $P(1,7)$. What is the equation of the circle?

I thought you could solve it by $R^2 = 3^2 + 4^2$, but that would just give a radius of 5 because there aren't any $x$ or $y$'s. Then I thought, should you then add a point $P(x,y)$? I'm not too sure, can someone give me hints?

  • $\begingroup$ Consult this $\endgroup$ – Sasha Sep 13 '12 at 21:06
  • $\begingroup$ If the coordinates of the centre of a circle is (a,b) and the radius of the circle is c, then the equation of the circle is given by: (x-a)^2 + (y-b)^2 = c^2 $\endgroup$ – Adam Rubinson Sep 13 '12 at 21:07

You correctly calculated the radius, now you just need to shift the circle into position. The general equation is

$r^2 = (x-x_c)^2 + (y-y_c)^2$

Where the center is $(x_c,y_c).$ So your equation becomes

$r^2 = (x+2)^2 + (y-3)^2$

You do not need to add an extra point, since a circle is already determined by its center and an arbitraty point on the circle.

  • $\begingroup$ Welcome to the site. If you enclose your $\LaTeX$ between dollar signs it looks nice. Double dollar signs give display mode. You can click on the time after "edited" to see what I did. $\endgroup$ – Ross Millikan Sep 13 '12 at 21:10
  • $\begingroup$ But why can't you just write it down as 5 = (x+2) + (y-3) ? I know that is a straight line, but, I got r which is 5, which means I can use (x+2) + (y-3) instead of (x+2)^2 + (y-3)^2 .. why doesn't it work out? $\endgroup$ – PhysicalEntity Sep 13 '12 at 21:10
  • $\begingroup$ $\sqrt{a^2 + b^2} \neq a + b$ Actually, $(a+b)^2 = a^2 + 2ab + b^2$ $\endgroup$ – Sidd Singal Sep 13 '12 at 21:12
  • $\begingroup$ Thanks Ross for editing my answer. Looks way better ;) $\endgroup$ – Alexander Körschgen Sep 13 '12 at 21:13
  • $\begingroup$ @ Sidd, of course, stupid mistake of mine. I get this now! $\endgroup$ – PhysicalEntity Sep 13 '12 at 21:14

The equation of a circle, centre $(p,q)$ and radius $r$, is

$$ (x-p)^2 + (y-q)^2 = r^2 \, . $$

Here $p$, $q$ and $r$ will be fixed numbers and the circle is the set of points $(x,y)$ for which the equation holds, for example: $(x-1)^2 + (y+2)^2 = 4$ is the equation of the circle with centre $(1,-2)$ and radius $2$.

In your case, the centre is $(-2,3)$. There are lots of circles centred at $(-2,3)$. They have equations $(x+2)^2 + (y-3)^2 = r^2$ where $r > 0$ is the radius of the circle. You also know that the circle passes through the point $(1,7)$. We can use this to find the radius. The circle $(x+2)^2 + (y-3)^2 = r^2$ passes through the point $(1,7)$ if and only if $(x,y) = (1,7)$ is a solution to the equation $(x+2)^2 + (y-3)^2 = r^2$. Let's put $(x,y) = (1,7)$ into the equation:

$$(1+2)^2 + (7-3)^2 = r^2 \iff 3^2 + 4^2 = r^2 \iff 25 = r^2 \, . $$

Since $r>0$, we have $r = 5$ (instead of $r = -5$). It follows that your circle has the equation

$$(x+2)^2 + (y-3)^2 = 25 \, . $$

It is the circle centred at $(-2,3)$ with radius $5$.


You correctly figured out the distance from the center of the circle to point P on the circle. You did this through the calculation


to obtain a radius of length 5. Now, from the definition of a circle, the distance from the center to each point $(x,y)$ on the circle is the same. This distance can be calculated in the exact same way.

$$[x-(-2)]^2+(y-3)^2=5^2\text{ or }(x+2)^2+(y-3)^2=25$$


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