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

The Correct Option is A

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:

\[ x^2 + y^2 - 65 = 0 \]

Answer 2

The problem involves finding the locus of point \(C(h, k)\) given that \(P(6, 1)\) is the orthocenter of triangle \(ABC\) with known vertices \(A(5, -2)\) and \(B(8, 3)\). The goal is to determine on which circle point \(C\) lies.

1. First, recall that the orthocenter is the intersection of the altitudes in a triangle. We know \(P\), the orthocenter, has coordinates \((6, 1)\).
2. To find the equation involving \(C\), we first need the slopes of the altitudes:
3. Calculate the slope of line \(AB\): \(\text{slope of } AB = \frac{3 - (-2)}{8 - 5} = \frac{5}{3}\).
4. The altitude from \(C\) to \(AB\) is perpendicular to \(AB\), so its slope will be the negative reciprocal of \(AB\)'s slope: \(-\frac{3}{5}\).
5. With the slope of the altitude through \(C\) and knowing it passes through the orthocenter \(P(6, 1)\), we can write its equation: 
   \(y - 1 = -\frac{3}{5}(x - 6)\) 
   Simplifying gives: \(5(y - 1) = -3(x - 6)\), \(\Rightarrow 5y - 5 = -3x + 18\), \(\Rightarrow 3x + 5y = 23\).
6. (Optional) Similarly, we could find another condition using another altitude through another vertex, but we already have enough to determine \(C\)'s nature.
7. To find the locus of \(C\), substitute known equations into each given circle equation. Since \(C\) satisfies \(3x + 5y = 23\), plug this into each circle:
   • \(x^2 + y^2 - 65 = 0\): Substitute \(y\) from the line equation: 
     Using \((3x + 5y = 23 \Rightarrow y = \frac{23 - 3x}{5})\), 
     \[ x^2 + \left(\frac{23 - 3x}{5}\right)^2 - 65 = 0 \] 
     Solving gives consistency with data for potential \(h, k\).
   • Check the other equations similarly, observing redundancy or unsuitability.
8. Through elimination and verification, conclude that the only consistent equation for all valid pairs \((h, k)\) for \(C\) is \(x^2 + y^2 - 65 = 0\).

Thus, the correct answer is that the point \(C\) lies on the circle: \(x^2 + y^2 - 65 = 0\).