Find the graph representing the equation $(x-2)^2+(y-3)^2+(x-2)(y-3)=0$

Question

Locus Find the graph representing the equation $(x-2)^2+(y-3)^2+(x-2)(y-3)=0$

After breaking it I am getting the discriminant of the second degree conic to be non zero and $h^2-ab<0$

So, it should be an ellipse.Am i going right?

Answer 1

Take $u=x-2$, $v=y-3$. This moves your origin to $(2,3)$. Now your equation becomes $$0=u^2+v^2+uv=\frac{3}{4}(u+v)^2+\frac{1}{4}(u-v)^2$$

Now multiply both sides by $4$ and take $a=u+v$, $b=u-v$. This rotates your axes. Now your equation becomes $$0=3a^2+b^2$$

This has only one solution, namely $a=b=0$. So you get just a single point, a degenerate conic.

Answer 2

Perhaps another way to see this is to measure polar angle from $ \ (2, \ 3 ) \ $ to points on the curve either by writing $ \ x \ - \ 2 \ = \ r \ \cos \theta \ \ , \ \ y \ - \ 3 \ = \ r \ \sin \theta \ $ , or shifting the origin of the coordinate system first as vadim123 does. The curve equation becomes

$$ r^2 \cos^2 \theta \ + \ r^2 \sin^2 \theta \ + \ r^2 \sin \theta \cos \theta \ = \ 0 \ \ \Rightarrow \ \ r^2 \ ( 1 \ + \ \frac{1}{2} \ \sin \ 2 \theta ) \ = \ 0 \ \ . $$

The factor in parentheses is always positive, so the only solution is $ \ r \ = \ 0 \ $ , a single point.