Finding a constant k such that a circle equation has center (3,1)

Question

Given that the equation of a circle is $x^2+y^2+8x+4y-2+k(x^2+y^2+x+y+3)=0$, for some constant k. Determine the value of $k$ such that the center of the circle will be at $(3,1)$.

I have tried a few different things, but I cannot seem to get a solution. I know the final formula should look like $(x-3)^2+(y-1)^2=R$, but how can I know what R is?

Answer 1

Equation can be formulated as

$$x^2+y^2+\frac{8+k}{2(k+1)}x+\frac{4+k}{2(k+1)}y+3k-2=0\\ \left(x+\frac {8+k}{2(k+1)}\right)^2+\left(y+\frac {4+k}{2(k+1)}\right)^2=\underbrace{-\frac {5k^2-10k-44}{2(k+1)^2}}_{R^2}$$ Putting $\color{red}{k=-2}$ gives $$(x-3)^2+(y-1)^2=2$$ i.e. $\color{red}{R=\sqrt2}$

Answer 2

Hint: Expand the terms in the formula you are given and your final formula and match them up. For example, in the formula you are given the terms involving $x$ are $(k+1)x^2+(8+k)x$. Remember that you can multiply one of them by a constant to make things match. You should be able to find values of $k,R$ that make them all match.

Answer 3

If $k=-1$ we can not get an equation of a circle.

Thus, $k\neq-1$ and we have $$(k+1)x^2+(k+1)y^2+(k+8)x+(k+4)y+3k-2=0$$ or $$x^2+y^2+\frac{k+8}{k+1}x+\frac{k+4}{k+1}y=\frac{2-3k}{k+1}$$ or $$\left(x+\frac{k+8}{2(k+1)}\right)^2+\left(y+\frac{k+4}{2(k+1)}\right)^2=\left(\frac{k+8}{2(k+1)}\right)^2+\left(\frac{k+4}{2(k+1)}\right)^2+\frac{2-3k}{k+1}.$$ Now, we need $\frac{k+8}{2(k+1)}=-3$ and $\frac{k+4}{2(k+1)}=-1$, which is impossible.

Answer 4

Since you already know what the center is supposed to be, expanding both equations and matching up coefficients, per Ross Millikan’s answer, is a straightforward way to approach this. Here are a couple more.

If the coefficient of $x^2+y^2$ in the equation of the circle is $1$, then you can read the coordinates of its center directly from the coefficients of the linear terms. You can find this correspondence for yourself by expanding $(x-h)^2+(y-k)^2=r^2$ and examining the coefficients of $x$ and $y$. Your equation expands into $$(k+1)(x^2+y^2)+(k+8)x+(k+4)y+\dots$$ We can divide by $k+1$ since this is no longer the equation of a circle if this is zero, so the circle’s center is at $x=-{k+8\over2(k+1)}$, $-{k+4\over2(k+1)}$. Set these coordinates equal to the known center and solve for $k$.

Alternatively, you can find the center of any conic (that has a center, that is) by differentiating its general-form equation and solving for the values of $x$ and $y$ at which both partial derivatives vanish. For your equation, this means solving the system $$\begin{align}2x+8+2kx+k &= 0 \\ 2y+4+2ky+k &= 0\end{align}$$ for $x$ and $y$, and then solving for $k$ as above.