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CUET Previous Year Questions (PYQs)

CUET Circle PYQ

CUET PYQ
List I List II
A. Dog : Rabies :: Mosquito : I. Bacteria
B. Amnesia : Memory :: Paralysis : II. Liver
C. Meningitis : Brain :: Cirrhosis : III. Movement
D. Influenza : Virus :: Typhoid : IV. Malaria






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CUET Previous Year PYQCUET CUET 2023 PYQ

Solution

Dog causes Rabies, Mosquito causes Malaria → A-IV
Amnesia affects Memory, Paralysis affects Movement → B-III
Meningitis affects Brain, Cirrhosis affects Liver → C-II
Influenza is caused by Virus, Typhoid is caused by Bacteria → D-I

Correct matching: A-IV, B-III, C-II, D-I
CUET PYQ
A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2 + y^2 = 16$ and $x^2 + y^2 = 1$. Then





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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Let the centre of circle $S$ be $(h,k)$ and radius $r$.

Orthogonality condition with circle $x^2+y^2=1$:
$h^2 + k^2 = r^2 + 1$

Orthogonality with $(x-1)^2+y^2=16$:
$(h-1)^2 + k^2 = r^2 + 16$

Subtracting:
$(h-1)^2 - h^2 = 15$
$h^2 - 2h +1 - h^2 = 15$
$-2h = 14 \Rightarrow h = -7$

Since circle passes through $(0,1)$:
$r^2 = (0+7)^2 + (1-k)^2$

Using $h^2+k^2=r^2+1$:
$49 + k^2 = r^2 + 1$

Solving gives $k=1$.

Centre of $S = (-7,1)$
CUET PYQ
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A:
If two circles intersect at two points, then the line joining their centres is perpendicular to the common chord.

Reason R:
The perpendicular bisectors of two chords of a circle intersect at its centre.
In the light of the above statements, choose the correct answer from the options given below:





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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Assertion A is a standard geometric property of intersecting circles ⇒ True.

Reason R is also true (property of chords in a circle), but it does not directly explain why the line joining centres is perpendicular to the common chord of two circles.
CUET PYQ
The center and radius for the circle $x^2 + y^2 +6x-4y +4 = 0$ respectively are: 
1. (2, 3) and 3 
2. (3, 2) and 8 
3. (2, -3) and 3 
4. (-3, 2) and 3





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CUET Previous Year PYQ CUET CUET 2025 PYQ

Solution

Solution

Equation: $$x^2 + y^2 + 6x - 4y + 4 = 0$$

Step 1: Group terms.
$$(x^2 + 6x) + (y^2 - 4y) + 4 = 0$$

Step 2: Complete the square.
- For $x^2 + 6x$: add and subtract $(\tfrac{6}{2})^2 = 9$
- For $y^2 - 4y$: add and subtract $(\tfrac{-4}{2})^2 = 4$

$$(x^2 + 6x + 9) + (y^2 - 4y + 4) + 4 - 9 - 4 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 - 9 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 = 9$$

Center: (-3, 2)

Radius: 3

Answer: Option 4

CUET PYQ
Which of the following option is false about circles whose equations
$ x^2 + y^2 - 10x - 10y + 41 = 0 $
and
$ x^2 + y^2 - 22x - 10y + 137 = 0 $
(A) circles have same center 
(B) circles have no meeting time 
(C) circles have only one meeting time 
(D) circles have only two meeting time





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CUET Previous Year PYQ CUET CUET MCA 2026 PYQ

Solution

CUET PYQ
The equation of a circle that passes through the points (3, 0) and (0, –2) and its lies on a line 2x + 3y = 3 then equation of the cicle is given by:





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CUET Previous Year PYQ CUET CUET 2024 PYQ

Solution


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