Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone please solve these 2 equations to get values of h and k? I know the values of h and k but not sure how to solved these equations to get h and k 's values

  1. $(20.01 - h)^2 + (17.94 - k)^2 = 285.27$
  2. $(3.25 - h)^2 + (15.81 - k)^2 = 285.27$

These equations are of a circle representing 2 points on circumferences. Answer is $h = 13.45$ $k = 2.36$

share|improve this question
Are you asking for a solution for $h$ and $k$ in terms of $r$? Please clarify. –  Alex Becker Jan 6 '12 at 6:31
The two equations are not enough to solve for $h$ and $k$ without knowing what $r$ is. There are infinitely many distinct circles that go through the points $(20.01,17.94)$ and $(3.25,15.81)$, and so the two equations, by themselves, cannot uniquely determine $h$ and $k$. –  Arturo Magidin Jan 6 '12 at 6:34
Ok here is the value of r^2 = 285.27 –  coure2011 Jan 6 '12 at 6:35
Are you sure there's only one solution for $(h, k)$? –  Sp3000 Jan 6 '12 at 7:08
@coure2011: There are two solutions for each value of $r$. –  Arturo Magidin Jan 6 '12 at 17:28
show 1 more comment

2 Answers

Let's discuss the idea behind trying to do this.

You have two points on a circle, $(a,b)$ and $(c,d)$. You know that the center of the circle, wherever it may be, must be equidistant to $(a,b)$ and $(c,d)$ (that is, the distance from the center to $(a,b)$ must be equal to the distance from the center to $(c,d)$).

However, any point that is equidistant to $(a,b)$ and $(c,d)$ will work as the center of some circle that goes through $(a,b)$ and $(c,d)$: if $(h,k)$ has distance $r$ to $(a,b)$ and $(c,d)$, then $$(x-h)^2 + (y-k)^2 = r^2$$ will be a circle with center $(h,k)$ that goes through $(a,b)$ and $(c,d)$.

What are all the points that are equidistant to $(a,b)$ and $(c,d)$? It is the line that is perpendicular to the segment that joins $(a,b)$ and $(c,d)$ and goes through the midpoint of that line; see for example here. It's known as the "perpendicular bisector".

The line segment from $(a,b)$ to $(c,d)$ is given by the parametric equations $$\begin{align*} x &= a + t(c-a),\ y &= b + t(d-b), \end{align*}\qquad $0\leq t\leq 1,$$ and the midpoint is given when $t=\frac{1}{2}$; that is, the midpoint is $$\left(\frac{a+c}{2}, \frac{b+d}{2}\right).$$ If $b=d$, then the segment is horizontal, so the perpendicular bisector is vertical with equation $x = \frac{a+c}{2}$. Any point on that line will be a center of a circle through those two points.

If $b\neq d$, then the slope of the perpendicular bisector is $m=-\frac{c-a}{d-b}$ (the negative reciprocal of the slope of the line through $(a,b)$ and $(c,d)$), so the equatio of the perpendicular bisector is $$ y = -\frac{c-a}{d-b}\left(x - \frac{a+c}{2}\right) + \frac{b+d}{2}.$$ Any point on that line is the center of a circle that goes through $(a,b)$ and $(c,d)$.

So there are infinitely many solutions to your two equations.

If you know the value of $r$, that reduces the possible solutions to $2$, except if one case: the only case where there is a unique solution is the case where the distance is exactly half the distance between $(a,b)$ and $(c,d)$; otherwise, you always have two solutions, that are symmetrically placed on both sides of the line segment through $(a,b)$ and $(c,d)$.

In the case you give, we have $(a,b) = (20.1, 17.94)$, $(c,d)=(3.25,15.81)$; so the center of any circle through those two points lies on the line $$ y = -\frac{3.25-20.1}{15.81-17.92}\left( x - \frac{23.35}{2}\right) + \frac{33.75}{2} = \frac{16.85}{2.11}\left(x - 11.625\right) + 16.875.$$ If you know that $r^2 = 285.27$, then this gives you two possible points on this line whose distance to $(a,b)$ and $(c,d)$ is $\sqrt{r}$. Just take $(x-a)^2 + (y-b)^2 = r^2$, plug in the values of $a$, $b$, $r^2$, substitute the value of $y$ for the expression on $x$ given above, and solve for $x$ (e.g., using the quadratic equation). That will give you the two values of $x$ that correspond to $h$, with the expression for $y$ giving the corresponding values of $k$.

share|improve this answer
add comment
  1. $(20.01 - h)^2 + (17.94 - k)^2 = 285.27$
  2. $(3.25 - h)^2 + (15.81 - k)^2 = 285.27$

From 1, we have:

$20.01^2-40.02h+h^2+ 17.94^2-35.88k+k^2=285.27.$

i.e. $40.02h+35.88k=h^2+k^2+436.9737\cdots (3)$

From $(2)$, we have:


i.e. $6.5h+31.62k=h^2+k^2-24.7514\cdots (4)$

From $(3),(4)$ we get:


i.e. $33.52h+4.26k=461.7251$

i.e. $h=13.7746152-0.127088305k\cdots(5)$

Plugging this value in, say $(3)$, we get:


i.e. $1.01615144k^2-34.295111k+75.4536236=0.$

Solving this, we get $k=31.38400797219823$ & $k=2.365991929391239$.

Corresponding to approximate value $k=2.36$, from $(5)$, we see that


i.e. $h=13.4746868$ i.e. $h=13.47$ approximately.

Thus, we have: $(h,k)=(2.36,13.47)$. Note that there exists another value for $k$ and hence for $h$, so the two circles intersect each other in two points, one of which is $(2.36,13.47)$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.