Question

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.

• A. $x^2 + y^2 - 65 = 0$
• B. $x^2 + y^2 - 74 = 0$
• C. $x^2 + y^2 - 61 = 0$
• D. $x^2 + y^2 - 52 = 0$

Answer 1

Answer: (A)

$x^2 + y^2 - 65 = 0$

Answer 2

To find the coordinates of $C$, we proceed with the slopes of sides and the equations of lines.

1. Slope of $AD$:

$\text{Slope of } AD = 3$

2. Slope of $BC$:

$\text{Slope of } BC = -\frac{1}{3}$

Equation of $BC$:

$3y + x - 17 = 0$

3. Slope of $BE$:

$\text{Slope of } BE = 1$

4. Slope of $AC$:

$\text{Slope of } AC = -1$

Equation of $AC$:

$x + y - 3 = 0$

Solving these equations, we find:

$\text{Point } C \text{ is } (-4, 7)$

Since $C$ lies on the circle, we have:

$x^2 + y^2 - 65 = 0$