CUET PG 2023 — Mathematics PYQ

Q: How do I solve CUET PG 2023 Mathematics questions?

Practice CUET PG 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of CUET PG Mathematics questions?

CUET PG Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are CUET PG previous year papers available for free?

Yes. Mathem Solvex provides free access to CUET PG previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

CUET PG 2023 — Mathematics PYQ

CUET PG | Mathematics | 2023

A circle S passes through the point (0, 1) and is orthogonal to the circles (𝑥−1)²+𝑦²=16 and 𝑥²+𝑦²=1. Then

(A) Radius of S is 8 (B) Radius of S is 7
(C)Centre of S is (-7, 1) (D)Centre of S is (-8. 1)

Choose the correct answer:

Explanation

Step 1: Point (0, 1) se guzarne ki condition

Sawaal ke anusar, circle $(0, 1)$ se guzarta hai, toh:

$0^2 + 1^2 + 2g(0) + 2f(1) + c = 0$

1 + 2f + c = 0 \quad \text{--- (Equation 1)}

Step 2: Pehle circle ke saath Orthogonality

Pehla circle hai: $(x-1)^2 + y^2 = 16 \implies x^2 + y^2 - 2x - 15 = 0$

Yahan $g_1 = -1, f_1 = 0, c_1 = -15$ .

Orthogonality ki condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ ka use karte hue:

$2(-1)(g) + 2(0)(f) = -15 + c$

$-2g = -15 + c$

c = 15 - 2g \quad \text{--- (Equation 2)}

Step 3: Dusre circle ke saath Orthogonality

Dusra circle hai: $x^2 + y^2 = 1 \implies x^2 + y^2 - 1 = 0$

Yahan $g_2 = 0, f_2 = 0, c_2 = -1$ .

Condition apply karne par:

$2(0)(g) + 2(0)(f) = -1 + c$

$0 = -1 + c$

c = 1 \quad \text{--- (Equation 3)}

Step 4: $g$ aur $f$ ki values nikalna

Equation 3 se humein mila $c = 1$ . Ab ise baki equations mein rakhte hain:

$g$ nikalne ke liye (Equation 2):

$1 = 15 - 2g \implies 2g = 14 \implies \mathbf{g = 7}$
$f$ nikalne ke liye (Equation 1):

$1 + 2f + 1 = 0 \implies 2f = -2 \implies \mathbf{f = -1}$

Step 5: Centre aur Radius

Circle $S$ ka centre $(-g, -f)$ hota hai:

\text{Centre} = (-7, 1)

Ab radius ( $r$ ) nikalte hain:

$r = \sqrt{g^2 + f^2 - c}$

$r = \sqrt{7^2 + (-1)^2 - 1}$

$r = \sqrt{49 + 1 - 1} = \sqrt{49}$

\text{Radius} = 7

Right Answer

Aapke options ke hisaab se:

(B) Radius of S is 7
(C) Centre of S is (-7, 1)