Animated Solution for Mathematics - Straight Lines: If $P(6,1)$ be the orthocentre of the triangle whose vertices are $A(5,-2)$, $B(8,3)$ and $C(h,k)$, then the point $C$ lies on the circle:

Welcome, fellow traveler on the path of JEE mastery. Today, we are not just solving a coordinate geometry problem; we are uncovering the hidden architecture of a triangle.

We are given two vertices, $A(5,-2)$ and $B(8,3)$, and the orthocenter $P(6,1)$. Our mission is to find the third vertex $C(h,k)$ and identify the circle upon which it resides.

Many students see this and immediately reach for the distance formula, hoping to set up a system of equations. But stop. Breathe. Geometry is not about brute force; it is about relationships.

The orthocenter is the meeting point of the altitudes. This is the key that unlocks the entire problem. If $P$ is the orthocenter, then the line $AP$ is the altitude from $A$ to $BC$, and $BP$ is the altitude from $B$ to $AC$. This means $AP\perp BC$ and $BP\perp AC$.

Let us begin by calculating the slope of the altitude $AP$. The slope formula, $m=\frac{y_2-y_1}{x_2-x_1}$, is our most trusted tool. Substituting the coordinates of $A(5,-2)$ and $P(6,1)$, we get:

Now, because $AP$ is perpendicular to $BC$, their slopes must satisfy the condition $m_{AP}\cdot m_{BC}=-1$. This is the "negative reciprocal" rule. Thus, the slope of side $BC$ is simply:

We repeat this elegant dance for the second altitude, $BP$. Using $B(8,3)$ and $P(6,1)$:

Since $BP\perp AC$, the slope of side $AC$ must be the negative reciprocal of $1$, which is $-1$. We have now mapped the slopes of the sides of our triangle. The geometry is beginning to reveal itself.

With the slopes in hand, we can now define the lines $BC$ and $AC$. We use the point-slope form, $y-y_1=m(x-x_1)$.

For line $BC$, passing through $B(8,3)$ with slope $m_{BC}=-\frac{1}{3}$:

Multiplying by $3$ and rearranging, we get:

For line $AC$, passing through $A(5,-2)$ with slope $m_{AC}=-1$:

Simplifying this, we get:

Now, we stand at the threshold of the solution. Vertex $C$ is the intersection of lines $BC$ and $AC$. We have a system of two linear equations:

Subtracting equation $(2)$ from equation $(1)$ is a moment of pure mathematical satisfaction:

Substituting $y=7$ back into equation $(2)$:

Our vertex $C$ is at $(-4,7)$.

Finally, we check the circle options. We need to find the equation $x^2+y^2+\cdots =0$ that $C(-4,7)$ satisfies. Calculating $x^2+y^2$ for our point:

Thus, the point $C$ lies on the circle $x^2+y^2-65=0$. We have navigated the geometry, applied the properties of the orthocenter, and arrived at the result with precision. This is the beauty of JEE mathematics—when you understand the underlying principles, the path to the answer becomes clear and inevitable.