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

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Q.
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 [2024]

1 $x^{2} + y^{2} - 65 = 0$

2 $x^{2} + y^{2} - 74 = 0$

3 $x^{2} + y^{2} - 52 = 0$

4 $x^{2} + y^{2} - 61 = 0$

Ans.
(1)

Slope of $A P = \frac{1 + 2}{6 - 5} = 3$

Slope of $B C = \frac{- 1}{Slope of A D}$

$= \frac{- 1}{Slope of A P}$

$= \frac{- 1}{3}$

Equation of line BC is given by $(y - 3) = \frac{- 1}{3} (x - 8)$

$\Rightarrow 3 y + x - 17 = 0$ ...(i)

Now, slope of $B P = \frac{1 - 3}{6 - 8} = 1$

Slope of $A C = - 1$

Equation of AC is $y + 2 = - 1 (x - 5)$

i.e., $y + x - 3 = 0$ ...(ii)

C is the point of intersection of AC and BC, then solving equation (i) and (ii), we get

$C = (- 4, 7), which lies on the circle x^{2} + y^{2} - 65 = 0 .$

Solution

Q.
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 [2024]

1 $x^{2} + y^{2} - 65 = 0$

2 $x^{2} + y^{2} - 74 = 0$

3 $x^{2} + y^{2} - 52 = 0$

4 $x^{2} + y^{2} - 61 = 0$

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Solution

Q. 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 [2024]

1 x2+y2-65=0

2 x2+y2-74=0

3 x2+y2-52=0

4 x2+y2-61=0

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Q.
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 [2024]

1 $x^{2} + y^{2} - 65 = 0$

2 $x^{2} + y^{2} - 74 = 0$

3 $x^{2} + y^{2} - 52 = 0$

4 $x^{2} + y^{2} - 61 = 0$