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MAH-CET 2025 Mathematics PYQ — A family of circles lies in the second quadrant and touches both … | Mathem Solvex | Mathem Solvex

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MAH-CET 2025 — Mathematics PYQ

MAH-CET | Mathematics | 2025

A family of circles lies in the second quadrant and touches both the x-axis and y-axis. The general form of the equation for such a circle is: $(x + a)^{2} + (y - a)^{2} =$ $a^{2},a>0.$ Which of the following represent the differential equation for this family of circles?

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Same Topic Questions

MAH-CET

A square OABC is inscribed in a quadrant OPBQ of a circle …MAH-CET

The lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and$\frac{x…

A.
$\frac{d y}{d x} + \frac{y}{x} = 0$
B.
$x^{2} + y^{2} + 2 a x - 2 a y = 0$
C.
$\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - \frac{1}{y} = 0$
D.
$\frac{d y}{d x} = - \frac{y}{x}$
(Correct Answer)

Correct Answer:
$\frac{d y}{d x} = - \frac{y}{x}$

Explanation

Step 1: Given Equation

$(x + a)^2 + (y - a)^2 = a^2$

Step 2: Differentiate w.r.t. $x$

$2(x + a) + 2(y - a) \frac{dy}{dx} = 0$

Step 3: $a$ ki value nikalna

$(x + a) + (y - a) \frac{dy}{dx} = 0$

$x + a + y \frac{dy}{dx} - a \frac{dy}{dx} = 0$

$a(1 - \frac{dy}{dx}) = -(x + y \frac{dy}{dx})$

$a = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx} - 1}$

Step 4: Simple Relation ( $\frac{dy}{dx} = -\frac{y}{x}$ )

Agar hum $a = 0$ maante hain (origin par circle), tab:

$x + y \frac{dy}{dx} = 0$

$y \frac{dy}{dx} = -x$

$\frac{dy}{dx} = -\frac{x}{y}$

Explanation

Step 1: Given Equation

MAH-CET
The incenter of the triangle formed with vertices
MAH-CET
Let

Choose the correct answer:

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