MereTeacher

JEE • NEET • FOUNDATION

JEE Previous Year Question

$The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is -$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
B $x^{2} + y^{2} - x - y = 0$
C $x^{2} + y^{2} + 2 x + 2 y - 7 = 0$
D $x^{2} + y^{2} + x + y - 2 = 0$

Correct answer: B

Solution

(b) Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius. (x - 1) (x - 0) + (y - 0) (y - 1) = 0 \Rightarrow x^{2} + y^{2} - x - y = 0

More Circle PYQs

$The circle on the chord x cos α + y sin α = p of the circle x^{2} + y^{2} = a^{2} as dia\dots$
$If the point (2, 0), (0, 1), (4, 5) and (0, c)\dots$
$If the circles x^{2} + y^{2} + 5 Kx + 2 y + K = 0 and 2 (x^{2} + y^{2}) + 2 Kx + 3 y - 1 = 0, (K ϵ R), intersect\dots$
$The two circles x^{2} + y^{2} = ax and x^{2} + y^{2} = c^{2} (c > 0) touch each other if:$
$If the points of intersection of the ellipses x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0 and 4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0 lie on a circl\dots$