Circles
Perpendicular Bisector — Sum of Abscissas
nta_pyq_2026_jan
Grade 11
Question:
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 points P is _____
Step-by-Step Solution
Key Concept: P $=(2\cos\theta,2\sin\theta)$ on circle. Q $=(\alpha,-5\alpha-2)$ on the line. Midpoint of PQ lies on $x-y+1=0$ and slope of PQ $=-1$. This gives two equations in $\theta$ and $\alpha$.
Sum of abscissas $=2/13$. $13\times(2/13)=2$.
Correct Answer: 2