JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q80

← JEE Main 2014 · 19 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 80 of 90

The equation of the circle described on the chord $3x+y+5=0$ of the circle $x^2+y^2=16$ as diameter is :

x^2+y^2+3x+y-11=0

x^2+y^2+3x+y+1=0

x^2+y^2+3x+y-2=0

x^2+y^2+3x+y-22=0

Answer and worked solution

Official answer: A

Take the family $x^2+y^2-16+\lambda(3x+y+5)=0$ . Its centre $\left(-\frac{3\lambda}2,-\frac\lambda2\right)$ must lie on the chord: $-\frac{9\lambda}2-\frac\lambda2+5=0$ , so $\lambda=1$ .
$x^2+y^2+3x+y-11=0.$

Topics:Family of circles

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 80