Mathematics · Circles

JEE Advanced 2018 — Paper 1 — Question 46

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then the mid-point of the line segment MN must lie on the curve

Option A:
$(x+y)^2 = 3xy$
Option B:
$x^{2/3} + y^{2/3} = 2^{4/3}$
Option C:
$x^2+y^2 = 2xy$
Option D:
$x^2+y^2 = x^2y^2$
Correct

Answer: D

Step-by-step solution

Circle equation: $x^2 + y^2 = 4$ . Let $P = (x_1, y_1)$ be on the circle with $x_1 > 0, y_1 > 0$ . Tangent at $P$ : $x x_1 + y y_1 = 4$ . X-intercept: set $y = 0$

$\Rightarrow M = (4/x_1, 0)$ . Y-intercept: set $x = 0$

$\Rightarrow N = (0, 4/y_1)$ . Midpoint $H$ of $MN$ : $(2/x_1, 2/y_1)$ . Write $(h, k) = (2/x_1, 2/y_1)$

$\Rightarrow x_1 = 2/h, y_1 = 2/k$ . Since $P$ lies on the circle: $(2/h)^2 + (2/k)^2 = 4$

$\Rightarrow 4/h^2 + 4/k^2 = 4$

$\Rightarrow 1/h^2 + 1/k^2 = 1$

$\Rightarrow k^2 + h^2 = h^2 k^2$ . Replacing $(h,k)$ with $(x,y)$ gives $x^2 + y^2 = x^2 y^2$ , which is option D.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Circles
Topic: Tangent & Normal , pair of tangents to circle , chord of contact

Let E 1 E 2 and F 1 F 2 be the chords of S passing through the point P 0 (1, 1) and parallel to the x-axis and the y-axis, respectively.…

Answer: D

Step-by-step solution

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