NIMCET 2027 Circle Previous Year Questions (PYQs) – Page 1 of 3
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Showing 22 questions
If the circles
$ x^2 + y^2 + 2x + 2ky + 6 = 0$
and
$x^2 + y^2 + 2ky + k = 0$
intersect orthogonally, then $k$ is:
1
2
3
4
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:
$k = 2 \;\text{or}\; -\dfrac{3}{2}$
Number of distinct solutions of
$ x^{2} = y^{2} $
and
$ (x - a)^{2} + y^{2} = 1 $
where $a$ is any real number:
1
2
3
4
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: $0,1,2,4$
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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2
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4
Solution
For the two circles $x^2+y^2=16$ and $x^2+y^2-2y=0$, there is/are
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2
3
4
Solution
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
1
2
3
4
Solution
The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangent to the same circle. The radius of the this circle is.
1
2
3
4
Solution
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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4
Solution
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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2
3
4
Solution
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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2
3
4
Solution
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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4
Solution
NIMCET 2027
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NIMCET 2027
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and More.
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