Q.Find the equation of the circle whose diameter has endpoints $(2,3)$ and $(-4,5)$.

Since the two given points are the ends of a diameter, the diameter-form equation $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ applies directly without needing to find the centre and radius first.

For diameter endpoints $(x_1,y_1)=(2,3)$ and $(x_2,y_2)=(-4,5)$, the equation is:

Expanding $(x-2)(x+4) = x^2+4x-2x-8 = x^2+2x-8$.

Expanding $(y-3)(y-5) = y^2-5y-3y+15 = y^2-8y+15$.

Adding: $x^2+2x-8+y^2-8y+15=0 \Rightarrow x^2+y^2+2x-8y+7=0$. …

A frequent error is writing $(x-x_2)$ as $(x-(-4))=(x-4)$ instead of correctly $(x+4)$ — carefully rewrite double negatives before expanding. Skipping the centre/midpoint cross-check i …