JEE Mains · Maths · STD 11 - 9. straight line
If P is a point on the circle \( x^{2}+y^{2}=4 \), Q is a point on the straight line \( 5x+y+2=0 \) and \( x-y+1=0 \) is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such point P is ........... .
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Mid point of PQ lies on \( x-y+1=0 \) \( \frac{2~cos~\theta+\alpha}{2}-\frac{2~sin~\theta-5\alpha-2}{2}+1=0 \) \( 2cos\theta+\alpha-2~sin\theta+5\alpha+2+2=0 \) \( cos\theta-sin\theta+3\alpha+2=0...(1) \) Slope of PQ is -1…
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