First slide

Definition of a circle

Question

The pole of the straight line $9 x + y - 28 = 0$ with respect to the circle $2 x^{2} + 2 y^{2} - 3 x + 5 y - 7 = 0,$ is

Moderate

A

$(3, 1)$
B

$(1, 3)$
C

$(3, - 1)$
D

$(- 3, 1)$

Solution

Let $(h, k)$ be the pole of the line $9 x + y - 28 = 0$ with respect to the circle $x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y - \frac{7}{2} = 0$ Then, the equation of the polar is
$\begin{matrix} h x + k y - \frac{3}{4} (x + h) + \frac{5}{4} (y + k) - \frac{7}{2} = 0 \\ \Rightarrow x (h - \frac{3}{4}) + y (k + \frac{5}{4}) - \frac{3}{4} h + \frac{5}{4} k - \frac{7}{2} = 0 \end{matrix}$
$\Rightarrow x (4 h - 3) + y (4 k + 5) - 3 h + 5 k - 14 = 0$
This equation and $9 x + y - 28 = 0$ represent the same line.
$\begin{array}{ll} ∴ \frac{4 h - 3}{9} = \frac{4 k + 5}{1} = \frac{- 3 h + 5 k - 14}{- 28} = λ (say) \\ \Rightarrow h = \frac{3 + 9 λ}{4}, k = \frac{λ - 5}{4}, - 3 h + 5 k - 14 = - 28 λ \\ \Rightarrow - 3 (\frac{3 + 9 λ}{4}) + 5 (\frac{λ - 5}{4}) - 14 = - 28 λ \end{array}$
$\begin{array}{l} \Rightarrow - 9 - 27 λ + 5 λ - 25 - 56 = - 112 λ \\ \Rightarrow - 22 λ - 90 = - 112 λ \\ \Rightarrow 90 λ = 90 \Rightarrow λ = 1 \end{array}$
Hence, the pole of the given line is $(3, - 1) .$

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