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The circle x^2+y^2=5 meets the parabola ...

The circle `x^2+y^2=5` meets the parabola `y^2=4x` at `P` and `Q` . Then the length `P Q` is equal to (a)2 (b) `2sqrt(2)` (c) 4 (d) none of these

A

2

B

`2sqrt(2)`

C

4

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
(3) Solving the circle `x^(2)+y^(2)=5` and the parabola `y^(2)=4x`, we have
`x^(2)+4x-5=0`
i.e., x=1 or x=-5 (not possible)
`:." "y^(2)=4ory=pm2`
Therefore, the points of intersection are P(1,2) and Q(1,-2).
Hence, PQ=4.
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