1.

Find the value of k if the points (4, 2) and (k, –3) are conjugate points with respect to the circle x2 + y2 – 5x + 8y + 6 = 0.

Answer»

Equation of the circle is x2 + y2 – 5x + 8y + 6 = 0 

Polar of P(4, 2) is 

x.4 + y.2 – (5/2) (x + 4) + 4 (y + 2) + 6 = 0 

8x + 4y – 5x – 20 + 8y + 16 + 12 = 0 

3x + 12y + 8 = 0 

P(4, 2), Q(k, –3) are conjugate points 

Polar of P passes through Q 

∴ 3k – 36 + 8 = 0 

3k = 28 ⇒ k = 28/3 .



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