Let P(3, 2) & ,Q (2 , 1) be two point and the straight lines PQ and QR are equally inclined to the circle $X^{2} + Y^{2} = 5$ then equation of the line QR

2 X - 11 Y = 15

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2 X + 11 Y = 15

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2 X + 11 Y + 15 = 0

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11 X - 2 Y = 15

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Solution

The correct option is B $2 X + 11 Y = 15$
PQ and QR making some angle with the tangle line at ( 2 , 1)
$∣ ∣ ∣ ∣ ∣ \frac{m - \frac{1}{2}}{1 + m \cdot \frac{1}{2}} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \frac{2 - \frac{1}{2}}{1 + 2 \cdot \frac{1}{2}} ∣ ∣ ∣ ∣ ∣$
$∣ ∣ ∣ \frac{2 m - 1}{m + 2} ∣ ∣ ∣ = \frac{3}{4} \Rightarrow 2 o r \frac{- 2}{11}$
Equation of QR is $Y - 1 = \frac{- 2}{11} (X - 2)$
$2 X + 11 y = 15$

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Let P(3, 2) & ,Q (2 , 1) be two point and the straight lines PQ and QR are equally inclined to the circle X2+Y2=5 then equation of the line QR

The correct option is B 2X+11Y=15PQ and QR making some angle with the tangle line at ( 2 , 1)∣∣ ∣ ∣∣m−121+m⋅12∣∣ ∣ ∣∣=∣∣ ∣ ∣∣2−121+2⋅12∣∣ ∣ ∣∣ ∣∣∣2m−1m+2∣∣∣=34⇒2or−211Equation of QR is Y−1=−211(X−2)2X+11y=15

Let P(3, 2) & ,Q (2 , 1) be two point and the straight lines PQ and QR are equally inclined to the circle $X^{2} + Y^{2} = 5$ then equation of the line QR