Coordinate Geometry: equation of a circle

Question

Find a value of k such that $k(x^2+y^2) + (y-2x+1)(y+2x+3)=0$ is a circle and hence find the center and radius of the resulting circle.

I've expanded the product on the right and end up with different coefficients (which means it's not a circle) for $x^2$ and $y^2$, namely (k-4) and (k+1) respectively.

Maybe I don't know enough algebra tricks, but I'm not sure what to do when the coefficients of $x^2$ and $y^2$ are different here.

This is my last ditch effort before accepting that the examination writers made another mistake, which for where I live is quite common.

Answer 1

Notice that

$$ (y-2x+1)(y+2x+3) = (y-2x+1)(y+2x+1) + 2 (y-2x+1) = ((y+1)^2-4x^2) + 2y-4x+2 = y^2+2y+1-4x^2+2y-4x+2=y^2+4y-4x^2-4x+3$$

Now, if we add $kx^2+ky^2$ to the equation we get

$$ ky^2 + y^2 + 4y + kx^2 - 4 x^2 - 4x + 3 $$

Answer 2

We have

$$ (k-4)x^2 + (k+1) y^2 + (-4)x + 4 y + 3 $$

If you look at the standard form of a conic section as:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F $$

and match coefficents we see the criterion for a ellipse becomes:

$$ B^2 - 4 A C < 0 \implies 0^2 - 4 (k-4)(k+1) < 0 $$

The further condition for circle is $B=0$ which is satisfied already and $A=C$. That sets $k-4=k+1$ which is impossible. So yes there is a mistake if you are told it is possible to be a circle.

However, you can proceed with asking for an ellipse.

$(k-4)(k+1)>0$ which gives $k>4$ or $k<-1$. All these $k$'s give ellipses.