NIMCET 2027 Circle Previous Year Questions (PYQs) – Page 2 of 3
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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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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
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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Solution
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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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,
$x^2-4x+4+y^2+2y+1=49$
$x^2+y^2-4x+2y-44=0$
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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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:
$\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$
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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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,
$\boxed{23}$
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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Solution
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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Solution
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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Solution
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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Solution
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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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 $= 2 + 33 + 33 = 68$
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 $= 2 + 33 + 33 = 68$
NIMCET 2027
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and More.
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