MCA NIMCET Circle Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Circle PYQ

Qus : 1 MCA 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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MCA NIMCET Previous Year PYQ MCA 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 MCA 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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MCA NIMCET Previous Year PYQ MCA 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 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 4 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 5 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 6 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 7 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 8 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 10 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 11 MCA 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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MCA NIMCET Previous Year PYQ MCA 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 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 13 MCA 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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MCA NIMCET Previous Year PYQ MCA 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 : 14 MCA 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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MCA NIMCET Previous Year PYQ MCA 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 : 15 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 16 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 17 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 18 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 19 MCA 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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MCA NIMCET Previous Year PYQ MCA 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 : 20 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 21 MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQ MCA 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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