Q.Find the circumcenter of the triangle whose vertices are $(1, 3)$, $(-3, 5)$, $(5, -1)$.

Since the angle in a semicircle subtending the hypotenuse (diameter) is $90°$, the perpendicularity condition on the vectors from the third vertex gives the locus equation.

Let the third vertex be $P(x,y)$. The hypotenuse endpoints are $A(0,6)$ and $B(6,0)$; since $AB$ is the hypotenuse, the angle at $P$ is $90°$, so $\vec{PA}\perp\vec{PB}$.

$\vec{PA} = (0-x,\ 6-y)$, $\qquad \vec{PB} = (6-x,\ 0-y)$

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Translate the ratio condition $PA:PB=2:3$ into $9\,PA^2=4\,PB^2$ using the distance formula, then simplify.

Let $P(x,y)$, $A(2,3)$, $B(2,-3)$. Given $\dfrac{PA}{PB}=\dfrac23 \implies 3\,PA=2\,PB \implies 9\,PA^2=4\,PB^2$.

$PA^2=(x-2)^2+(y-3)^2,\qquad PB^2=(x-2)^2+(y+3)^2$

$9\left[(x-2)^2+(y-3)^2\right] = 4\left[(x-2)^2+(y+3)^2\right]$

$5(x-2)^2 + \left[9(y-3)^2-4(y+3)^2\right]=0$

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With the right angle at $P(x,y)$, $PA\perp PB$, so $(x-0)(x-6)+(y-6)(y-0)=0$.

Let the third vertex be $P(x,y)$; the endpoints of the hypotenuse are $A(0,6)$ and $B(6,0)$. Since the right angle is at $P$, $\overline{PA}\perp\overline{PB}$, i.e. their dot product is zero:

$(x-0)(x-6)+(y-6)(y-0)=0$.

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Since the angle at the third vertex is $90^\circ$, the perpendicularity condition on the two legs gives the locus $x^2+y^2-4x-4y=0$ — the circle with $AB$ as diameter.

Concept

If $A(4,0)$ and $B(0,4)$ are the ends of the hypotenuse of a right triangle with the right angle at the third vertex $P(x,y)$, then $PA\perp PB$, so the product of their slopes is $-1$.

Step 1: Set up the perpendicularity condition

Slope of $PA = \dfrac{y-0}{x-4}$, slope of $PB = \dfrac{y-4}{x-0}$.

$\dfrac{y}{x-4}\cdot\dfrac{y-4}{x} = -1$

Step 2: Simplify

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