Circle 2 Question 1

1. If the circles x2+y2+5Kx+2y+K=0 and 2(x2+y2)+2Kx+3y1=0,(KR), intersect at the points P and Q, then the line 4x+5yK=0 passes through P and Q, for

(2019 Main, 10 April I)

(a) no values of K

(b) exactly one value of K

(c) exactly two values of K

(d) infinitely many values of K

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Answer:

Correct Answer: 1. (a)

Solution:

  1. Equation of given circles

x2+y2+5Kx+2y+K=0 and 2(x2+y2)+2Kx+3y1=0x2+y2+Kx+32y12=0

On subtracting Eq. (ii) from Eq. (i), we get

4Kx+12y+K+12=08Kx+y+(2K+1)=0

[ if S1=0 and S2=0 be two circles, then their common chord is given by S1S2=0.]

Eq. (iii) represents equation of common chord as it is given that circles (i) and (ii) intersects each other at points P and Q.

Since, line 4x+5yK=0 passes through point P and Q.

8K4=15=2K+1K

K=110 [equating first and second terms]

and K=10K+5

[equating second and third terms]

11K+5=0K=511

110511, so there is no such value of K, for which line 4x+5yK=0 passes through points P and Q.



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