MCQ

MCQ 11 Mark

If the equation of a circle is $\lambda x^2+(2 \lambda-3) y^2-4 x+$ $6 y-1=0$, then coordinates of centre are

A
$\left(\frac{4}{3},-1\right)$
✓
$\left(\frac{2}{3},-1\right)$
C
$\left(-\frac{2}{3}, 1\right)$
D
$\left(\frac{2}{3}, 1\right)$

Answer

Correct option: B.

$\left(\frac{2}{3},-1\right)$

(B) $\left(\frac{2}{3},-1\right)$
Explanation : If $\lambda x^2+(2 \lambda-3) y^2-4 x+6 y-1=0$
represents a circle, then coefficient of $x^2=$ coefficient of $y^2$
i.e., $\lambda=2 \lambda-3$
or $\lambda=3$
Now, the given equation can be written as$
3 x^2+3 y^2-4 x+6 y-1=0
$
or, $x^2+y^2-\frac{4}{3} x+2 y-\frac{1}{3}=0$
On comparing the above equation with $x^2+y^2+$$2 g x+2 f y+c=0$, we get
$2 g=-\frac{4}{3} \text { and } 2 f=2$
or $g=-\frac{2}{3} \text { and } f=1$
Hence, coordinates of the centre are $(-g,-f)$ i.e., $\left(\frac{2}{3},-1\right)$

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MCQ 21 Mark

The position of the point $(5,7)$ with respect to the circle, $x^2+y^2=100$ is

✓
inside the circle
B
outside the circle
C
lie on the circle
D
None of these

Answer

Correct option: A.

inside the circle

(A) inside the circle
Explanation : Here, $h^2+k^2=(5)^2+(7)^2$
$
\begin{array}{l}
=25+49 \\
=74<a^2=100
\end{array}
$
$\therefore$ Point $(5,7)$ is inside the circle.

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MCQ 31 Mark

The value of $\lambda$ for which the equation $2\left(x^2+y^2\right)-$ $6 x+8 y+\lambda=0$ represents a point circle is

A
$\frac{25}{4}$
B
$\frac{4}{25}$
✓
$\frac{25}{2}$
D
$\frac{2}{25}$

Answer

Correct option: C.

$\frac{25}{2}$

(C) $\frac{25}{2}$
Explanation : Given equation of circle is
$2\left(x^2+y^2\right)-6 x+8 y+\lambda=0$
or, $\left(x^2+y^2\right)-3 x+4 y+\frac{\lambda}{2}=0\quad \ldots(i) $
If the radius of the circle is zero, then the circle is called a point circle. Thus, eq. (i) will represents a point circle, if
$
\sqrt{\left(\frac{3}{2}\right)^2+(-2)^2-\frac{\lambda}{2}}=0\left[\because r=\sqrt{g^2+f^2-c}\right]
$
$\begin{array}{ll}\Rightarrow & \frac{\lambda}{2}=\frac{25}{4} \\ \Rightarrow & \lambda=\frac{25}{2}\end{array}$

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MCQ 41 Mark

If one end of a diameter of the circle $x^2+y^2-4 x-$ $6 y+11=0$ is $(3,4)$, then the coordinates of other end of the diameter is

✓
$(1,2)$
B
$(2,1)$
C
$(-1,2)$
D
$(2,-1)$

Answer

Correct option: A.

$(1,2)$

(A) $(1,2)$
Explanation : The given equation of circle is $x^2+y^2-4 x-6 y+11=0$ Its centre is $C(2,3)$. Let $A(3,4)$ and $B(\alpha, \beta)$ be the ends of the diameter, then $C(2,3)$ is the mid-point of the segment $A B$.

$\therefore \quad \frac{3+\alpha}{2}=2$
$\begin{array}{ll}\text { and } \frac{4+\beta}{2}=3 \\ \Rightarrow \alpha=1 \text { and } \beta=2\end{array}$
Hence, the coordinates of the other end of the diameter are $(1,2)$.

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MCQ 51 Mark

The equation of the circle which touches $x$-axis and whose centre is $(1,2)$ is

A
$x^2+y^2-2 x+4 y+1=0$
B
$x^2+y^2+2 x-4 y+1=0$
✓
$x^2+y^2-2 x-4 y+1=0$
D
$x^2+y^2-2 x-4 y-1=0$

Answer

Correct option: C.

$x^2+y^2-2 x-4 y+1=0$

(C) $x^2+y^2-2 x-4 y+1=0$
Explanation : Given, Centre of the circle is $C(1,2)$. As, the circle touches the $x$-axis, its radius $=$ Perpendicular distance from centre $(1,2)$ to the $x$-axis

$=$ ordinate of point $C=2$
$\therefore$ The equation of circle is $(x-1)^2+(y-2)^2=2^2$
i.e., $x^2+y^2-2 x-4 y+1=0$.

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MCQ 61 Mark

The equation of the circle with centre $(2,2)$ and which passes through the point $(4,5)$ is

A
$x^2+y^2-4 x-4 y=5$
B
$x^2+y^2+4 x+4 y=5$
C
$x^2+y^2-4 x+4 y=5$
D
$x^2+y^2+4 x-4 y=5$

Answer

(A) $x^2+y^2-4 x-4 y=5$
Explanation:The centre of the circle is $C(2,2)$ and it passes through point $P(4,5)$.

$\begin{array}{l}\therefore \text { Radius of circle }=C P \\ \quad=\sqrt{(4-2)^2+(5-2)^2} \\ \quad=\sqrt{4+9}=\sqrt{13}\end{array}$
$\therefore$ The equation of the circle is
$(x-2)^2+(y-2)^2=(\sqrt{13})^2$
$\Rightarrow\left(x^2-4 x+4\right)+\left(y^2-4 y+4\right)=13$
$\Rightarrow x^2+y^2-4 x-4 y=13-8$
$\Rightarrow x^2+y^2-4 x-4 y=5$

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MCQ 71 Mark

The equation of the circle whose centre is $(-2,5)$ and radius 3 will be

A
$x^2+y^2+4 x+10 y+20=0$
✓
$x^2+y^2+4 x-10 y+20=0$
C
$x^2+y^2+4 x+10 y-20=0$
D
$x^2+y^2-4 x-10 y+20=0$

Answer

Correct option: B.

$x^2+y^2+4 x-10 y+20=0$

(B) $x^2+y^2+4 x-10 y+20=0$
Explanation :The equation of circle whose centre is $(-2,5)$ and radius is 3 , is given by
$\begin{array}{l}(x+2)^2+(y-5)^2=3^2 \\ \Rightarrow\left(x^2+4 x+4\right)+\left(y^2-10 y+25\right)=9 \\ \Rightarrow x^2+y^2+4 x-10 y+29-9=0 \\ \Rightarrow x^2+y^2+4 x-10 y+20=0\end{array}$

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MCQ 81 Mark

The equation of circle which passes through the origin and whose centre is $(3,4)$ will be :

A
$x^2+y^2-3 x-4 y=0$
B
$x^2+y^2+3 x+4 y=0$
C
$x^2+y^2+6 x+8 y=0$
✓
$x^2+y^2-6 x-8 y=0$

Answer

Correct option: D.

$x^2+y^2-6 x-8 y=0$

(D) $x^2+y^2-6 x-8 y=0$
Explanation : If the origin be on the circumference of the circle and coordinates of the centre $C$ be $(3,4)$, then from the adjoining figure, it is clear that

$O C^2=O M^2+C M^2$
i.e., $r^2=3^2+4^2=9+16=25$
Thus, the equation of required circle with centre $(3,4)$ and radius 25 is given by
$(x-3)^2+(y-4)^2=25$
$\Rightarrow \quad\left(x^2-6 x+9\right)+\left(y^2-8 y+16\right)=25$
$\Rightarrow \quad x^2+y^2-6 x-8 y=0$

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Applied Maths MCQ

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