NIMCET Circle Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Circle PYQ

Qus : 1 NIMCET PYQ

If the circles $ x^2 + y^2 + 2x + 2ky + 6 = 0$ and $x^2 + y^2 + 2ky + k = 0$ intersect orthogonally, then $k$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

Two circles: $C_1:\; x^2 + y^2 + 2x + 2ky + 6 = 0$

$C_2:\; x^2 + y^2 + 2ky + k = 0$

Orthogonality condition for circles

$x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$

and

$x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ is:

$2(g_1g_2 + f_1f_2) = c_1 + c_2$

From $C_1$: $g_1 = 1,\; f_1 = k,\; c_1 = 6$

From $C_2$: $g_2 = 0,\; f_2 = k,\; c_2 = k$

Apply formula:

$2(1\cdot 0 + k\cdot k) = 6 + k$

$2k^2 = 6 + k$

$2k^2 - k - 6 = 0$

$(2k + 3)(k - 2) = 0$

So: $k = 2 \;\text{or}\; -\dfrac{3}{2}$

Qus : 2 NIMCET PYQ

Number of distinct solutions of $ x^{2} = y^{2} $ and $ (x - a)^{2} + y^{2} = 1 $ where $a$ is any real number:

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

$ x^{2} = y^{2} \Rightarrow x = \pm y $ Intersection with a circle shifted by $a$ gives variable counts depending on $a$. Possible solution counts: $0,1,2,4$

Qus : 3 NIMCET PYQ

The equation of the circle passing through the point (4,6) and whose diameters are along x + 2y - 5 =0 and 3x - y - 1=0 is

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 4 NIMCET PYQ

For the two circles $x^2+y^2=16$ and $x^2+y^2-2y=0$, there is/are

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 5 NIMCET PYQ

If a twelve sided regular polygon is inscribed in a circle of radius 3 centimeters, then the length of each side of the polygon is

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Qus : 6 NIMCET PYQ

The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangent to the same circle. The radius of the this circle is.

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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution

Qus : 7 NIMCET PYQ

The equation of a circle with diameters are 2x – 3y + 12 = 0 and x + 4y – 5 = 0 and area of 154 sq. units is

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Qus : 8 NIMCET PYQ

Find the equation of the circle which passes through (–1, 1) and (2, 1), and having centre on the line x + 2y + 3 = 0 .

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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 NIMCET PYQ

The number of common tangents to the circle $x^2+y^2=4$ and $x^2+y^2-6x-8y=24$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution

Qus : 10 NIMCET PYQ

The point of intersection os circle $x^2+y^2+10x-12y+51=0$ and the line $3y+x=3$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 11 NIMCET PYQ

The circle $x^2 + y^2 = 9$ is contained in the circle $x^2 + y^2 - 6x - 8y + 25 = c^2$ if

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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

Given circle 1: center (0,0), radius 3 Circle 2 (rewrite): (x-3)² + (y-4)² = c² Distance between centers = √(3² + 4²) = 5 Condition: larger radius ≥ smaller radius + distance c ≥ 3 + 5 = 8 Given options → smallest suitable c is 10

Qus : 12 NIMCET PYQ

The circles whose equations are $x^2+y^2+c^2=2ax$ and $x^2+y^2+x^2-2by=0$ will touch one another externally, if

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 13 NIMCET PYQ

Segments of the lines $2x+3y=1$ and $4x-3y=11$ are diameters of a circle of area $153.94$ square units. Then, the equation of this circle with integer radius is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

Since the given lines are diameters of the circle, both lines pass through the centre of the circle.

So, the centre is the intersection point of

$2x+3y=1$ .....$(1)$

$4x-3y=11$ .....$(2)$

Adding $(1)$ and $(2)$,

$6x=12$

$x=2$

Put $x=2$ in $(1)$:

$2(2)+3y=1$

$4+3y=1$

$3y=-3$

$y=-1$

So, centre of the circle is

$(2,-1)$

Now, area of circle is

$\pi r^2=153.94$

Since $153.94\approx 49\pi$,

$r^2=49$

$r=7$

Equation of circle is

$(x-2)^2+(y+1)^2=7^2$

$(x-2)^2+(y+1)^2=49$

Expanding,

$x^2-4x+4+y^2+2y+1=49$

$x^2+y^2-4x+2y-44=0$

Qus : 14 NIMCET PYQ

A circle with its center in the first quadrant touches both the coordinate axes and the line x-y-2=0. Then the area of the circle is

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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

A circle touching both coordinate axes has center $(r, r)$ and radius $r$.

It also touches the line $x - y - 2 = 0$.

So the distance from $(r, r)$ to the line equals $r$: $\dfrac{|r - r - 2|}{\sqrt{1^{2} + (-1)^{2}}} = r$

$\dfrac{2}{\sqrt{2}} = r$

$r = \sqrt{2}$

Area of the circle: $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$

Qus : 15 NIMCET PYQ

The circle $x^2 + y^2+ \alpha x+ \beta y+ \gamma=0$ is the image of the circle $x^2 + y^2- 6x- 10y+ 30=0$ across the line 3x + y = 2. The value of $[\alpha+ \beta+ \gamma]$ is (where [.] represents the floor function.)

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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

Given circle:

$x^{2}+y^{2}-6x-10y+30=0$

Center and radius: $C(3,5), \quad r=2$

We reflect $C(3,5)$ across the line $3x + y - 2 = 0$.

Compute reflection factor: $k = \dfrac{2(3\cdot 3 + 5 - 2)}{3^{2} + 1^{2}} = \dfrac{24}{10} = \dfrac{12}{5}$

Reflected center:

$x' = 3 - 3k = 3 - \dfrac{36}{5} = -\dfrac{21}{5}$

$y' = 5 - k = 5 - \dfrac{12}{5} = \dfrac{13}{5}$

So new center is:

$C'\left(-\dfrac{21}{5},\dfrac{13}{5}\right)$

The image circle is $x^{2}+y^{2}+\alpha x+\beta y+\gamma = 0$

Using center formula: $-\dfrac{\alpha}{2} = -\dfrac{21}{5} \Rightarrow \alpha = \dfrac{42}{5}$

$-\dfrac{\beta}{2} = \dfrac{13}{5} \Rightarrow \beta = -\dfrac{26}{5}$

Using radius $r=2$: $\left(-\dfrac{21}{5}\right)^{2} + \left(\dfrac{13}{5}\right)^{2} - \gamma = 4$

$\dfrac{610}{25} - \gamma = 4$

$\gamma = \dfrac{102}{5}$

Now compute:

$\alpha + \beta + \gamma = \dfrac{42}{5} - \dfrac{26}{5} + \dfrac{102}{5} = \dfrac{118}{5} = 23.6$

Thus, $\boxed{23}$

Qus : 16 NIMCET PYQ

A line passing through (4, 2) meets the x and y-axis at P and Q respectively. If O is the origin, then the locus of the centre of the circumcircle of ΔOPQ is -

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 17 NIMCET PYQ

Find the number of point(s) of intersection of the ellipse $\dfrac{x^2}{4}+\dfrac{(y-1)^2}{9}=1$ and the circle x² + y² = 4

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NIMCET Previous Year PYQ NIMCET NIMCET 2020 PYQ

Solution

Qus : 18 NIMCET PYQ

A circle touches the X-axis and also touches another circle with centre at (0, 3) and radius 2. Then the locus of the centre of the first circle is

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 19 NIMCET PYQ

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}$and having it centre at (0, 3) is

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 20 NIMCET PYQ

There are n equally spaced points 1,2,...,n marked on the circumference of a circle. If the point 15 is directly opposite to the point 49, then the total number of points is

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Given: $n$ equally spaced points $1, 2, \ldots, n$ marked on circumference of a circle
Point $15$ is directly opposite to point $49$
Since the points are equally spaced on a circle, two points are directly opposite if they are diametrically opposite
$\therefore$ The number of points between $15$ and $49$ on each semicircle must be equal
Points between $15$ and $49$ (going from $15$ to $49$):
$16, 17, 18, \ldots, 48$
Number of points $= 48 - 16 + 1 = 33$
Since both semicircles must have equal number of points between the opposite points:
Points on other semicircle (from $49$ to $15$) must also be $33$
$\therefore$ Total number of points $= 2 + 33 + 33 = 68$

Qus : 21 NIMCET PYQ

If two circles $x^{2}+y^{2}+2gx+2fy=0$ and $x^{2}+y^{2}+2g'x+2f'y=0$ touch each other then whichof the following is true?

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 22 NIMCET PYQ

. Two common tangents to the circles $x^2 + y^2 = 2a^2$ and parabola $y^2 = 8ax$ are

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Given circle: $x^2 + y^2 = 2a^2$
Given parabola: $y^2 = 8ax$
**Equation of tangent to parabola:**
For parabola $y^2 = 8ax$, tangent is:
$y = mx + \dfrac{2a}{m}$ ...(i)
**Condition for tangent to circle:**
For line $y = mx + \dfrac{2a}{m}$ to be tangent to circle $x^2 + y^2 = 2a^2$,
distance from centre $(0,0)$ = radius $= \sqrt{2}\ a$
$\Rightarrow \dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{1 + m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2a}{|m|\sqrt{1+m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2}{|m|\sqrt{1+m^2}} = \sqrt{2}$
$\Rightarrow |m|\sqrt{1+m^2} = \sqrt{2}$
Squaring both sides:
$\Rightarrow m^2(1+m^2) = 2$
$\Rightarrow m^4 + m^2 - 2 = 0$
$\Rightarrow (m^2 + 2)(m^2 - 1) = 0$
$\Rightarrow m^2 = 1 \quad (\because m^2 = -2 \text{ is not possible})$
$\Rightarrow m = \pm 1$
**Substituting in (i):**
When $m = 1$: $y = x + 2a$
When $m = -1$: $y = -x - 2a$
$\therefore$ The two common tangents are:
$\therefore \boxed{y = x + 2a \quad \text{and} \quad y = -(x + 2a)}$

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