JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q4

← JEE Main 2022 · 27 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 4 of 90

Let $\dfrac{dy}{dx}=\dfrac{ax-by+a}{bx+cy+a}$ , where $a,b,c$ are constants, represent a circle passing through the point $(2,5)$ . Then the shortest distance of the point $(11,6)$ from this circle is :

Answer and worked solution

Official answer: B

Circle condition. Cross-multiplying, $a\,x\,dx+a\,dx-b(y\,dx+x\,dy)-c\,y\,dy-a\,dy=0$ . A circle has no $xy$ term, so $b=0$ , and equal $x^2,y^2$ coefficients need $c=-a$ .

Circle. $x^2+y^2+2x-2y=k$ ; through $(2,5)$ : $k=23$ . Centre $(-1,1)$ , radius $\sqrt{25}=5$ .

Shortest distance from $(11,6)$ : $\sqrt{144+25}-5=13-5=8$ .

(B)

Topics:Formation and applications Equation of a circle

← Previous question Next question →

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 4