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.
- $x^2 + y^2 - 65 = 0$
- $x^2 + y^2 - 74 = 0$
- $x^2 + y^2 - 61 = 0$
- $x^2 + y^2 - 52 = 0$
The Correct Option is A
Solution and Explanation
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:
\[ x^2 + y^2 - 65 = 0 \]
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