is this equation solvable?

Question

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 $h = 13.45$ $k = 2.36$

Answer 1

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:

$3.25^2-6.5h+h^2+15.81^2-31.62k+k^2=285.27$.

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

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

$40.02h+35.88k-436.9737=6.5h+31.62k+24.7514$

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:

$40.02(13.7746152-0.127088305k)+35.88k=(13.7746152-0.127088305k)^2+k^2+436.9737$

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

$h=13.7746152-0.127088305\times2.36$

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 $(2.36,13.47)$.

Answer 2

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 $k$.