Fill in the blanks.

Question 11 Mark

Circle $x^2+y^2=9$ and $x^2+y^2+8 y+c=0$ will touch externally if $c=$ _________________ .

Answer

15 , because
Radius of circle $x^2+y^2=9$ is $r_1=3$
and radius of circle $x^2+y^2+8 y+c=0$ is
$
\begin{aligned}
r_2 & =\sqrt{g^2+f^2-c}=\sqrt{0^2+(4)^2-c} \\
& =\sqrt{16-c}
\end{aligned}
$
$\left[\because\right.$ On comparing equation $x^2+y^2+8 y+c=0$ with $x^2+y^2+2 g x+2 f y+c=0$, we get, $g=0$, $f=4$ and $c=c$ ]
Since, given circles touch externally
distance between their centres, $d=r_1+r_2\quad \ldots(i) $
Now, centre of $I ^{\text {st }}$ circle is $(0,0)$ and centre of $2^{\text {nd }}$ circle is $(-g,-f)$ i.e., $(0,-4)$
$\therefore d=\sqrt{(0-0)^2+\{0-(-4)\}^2}=\sqrt{16}=4$
Hence, from eq.(i), we get
$
\begin{array}{l}
4=3+\sqrt{16-c} \\
\Rightarrow \sqrt{16-c}=1 \\
\Rightarrow 16-c=1 \\
\Rightarrow c=15 \text {. }
\end{array}
$

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Question 21 Mark

Equation $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$ will represent a circle, if _________________ .

Answer

$a=b$ and $h=0$, because
General equation of second degree $a x^2+b y^2+2 h x y$ $+2 g x+2 f y+c=0$ will represent a circle, when
(i) coefficient of $x^2=$ coefficient of $y^2$ i.e., $a=b$
(ii) coefficient of $x y$ will be zero i.e., $2 h=0$ or $h=0$.

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Question 31 Mark

The straight line $y=m x+c$ will touch the circle $x^2+y^2=a^2$, if $c=$ _________________ .

Answer

$\pm a \sqrt{1+m^2}$, because
If the line touches the circle, there must be only one point which would satisfy both the equations
Hence,
$x^2+y^2=a^2$
$\begin{array}{l} \Rightarrow x^2+(m x+c)^2=a^2 \\ {[\because \text { putting } y=m x+c] }\end{array}$
$\Rightarrow x^2\left(1+m^2\right)+2 m c x+c^2=a^2$
$\Rightarrow x^2\left(1+m^2\right)+2 m c x+c^2-a^2=0$
The line touches the circle. Hence only one value of $x$ is possible. Therefore,
Discriminant $=0$
$\Rightarrow 4 m^2 c^2-4\left(1+m^2\right)\left(c^2-a^2\right)=0$
$\begin{array}{lrl}\Rightarrow m^2 c^2-c^2-m^2 c^2+a^2+m^2 a^2=0 \\ \Rightarrow a^2+a^2 m^2-c^2 =0 \\ \Rightarrow c^2 =a^2\left(1+m^2\right) \\ \Rightarrow c = \pm a \sqrt{1+m^2} .\end{array}$

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Question 41 Mark

If the circle $x^2+y^2-3 x+ k y-5=0$ and $4 x^2+4 y^2$ $-12 x-y-9=0$ be concentric, then $k=$ _________________ .

Answer

$-\frac{1}{4}$, because
Given equations of circles are
$
x^2+y^2-3 x+k y-5=0\quad \ldots(i)
$
$\therefore$ Centre of circle (i) is $\left(\frac{3}{2},-\frac{k}{2}\right)$
and $4 x^2+4 y^2-12 x-y-9=0\quad \ldots(ii) $
$\therefore$ Centre of circle (ii) is $\left(\frac{3}{2}, \frac{1}{8}\right)$
Since, the circles are concentric, therefore,
$
\frac{-k}{2}=\frac{1}{8} \Rightarrow k=-\frac{1}{4}
$

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Question 51 Mark

The value of ' $p$ ' so that the equation $x^2+y^2-2 p x$ $+4 y-12=0$ may represent a circle of radius 5 units is _________________ .

Answer

3 or -3 , because
Given equation of the circle is
$x^2+y^2-2 p x+4 y-12=0$
$\begin{array}{l}\Rightarrow \left(x^2-2 p x+p^2\right)+\left(y^2+4 y+4\right)-p^2-4-12=0 \\ \Rightarrow (x-p)^2+(y+2)^2=p^2+16 \\ \Rightarrow (x-p)^2+\{y-(-2)\}^2=\left(\sqrt{p^2+16}\right)^2\end{array}$
$\therefore$ The centre of the circle is $(p,-2)$ and radius $\sqrt{p^2+16}$ units.
According to the question,
$\sqrt{p^2+16}=5$
$\begin{aligned} \Rightarrow & p^2+16 =25 \\ \Rightarrow & p^2 =9 \\ \Rightarrow & p = \pm 3 .\end{aligned}$

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Question 61 Mark

The equation of the circle whose centre is $(-a,-b)$ and whose radius is $\sqrt{a^2+b^2}$ is _________________ .

Answer

$x^2+y^2+2 a x+2 b y=0$, because
Here, centre is $(-a,-b)$ and radius $=\sqrt{a^2+b^2}$
$\therefore$ The equation of circle is
$
\begin{array}{l}
{[x-(-a)]^2+[y-(-b)]^2=\left(\sqrt{a^2+b^2}\right)^2} \\
\Rightarrow (x+a)^2+(y+b)^2=a^2+b^2 \\
\Rightarrow \left(x^2+2 a x+a^2\right)+\left(y^2+2 b y+b^2\right)=a^2+b^2 \\
\Rightarrow x^2+y^2+2 a x+2 b y=0 \text {. }
\end{array}
$

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Question 71 Mark

Parametric equations of circles $x^2+y^2=4$ are _________________ .

Answer

$2 \cos t$ and $2 \sin t$, because
We know that, the parametric equations of the circle
$\begin{array}{l}x^2+y^2=r^2 \text { are } \\ x=r \cos t, y=r \sin t, 0 \leq t \leq 2 \pi\end{array}$
The given circle is comparable with $x^2+y^2=r^2$, here $r=2$.
Therefore, the parametric equations of the given circle $x^2+y^2=4$ are
$x=2 \cos t \text { and } y=2 \sin t, 0 \leq t \leq 2 \pi .$

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Question 81 Mark

The equation of the circle when end points of a diameter are $A(-2,3)$ and $B(3,-5)$ is _________________ .

Answer

$x^2+y^2-x+2 y-21=0$, because
We know that equation of the circle when the end points of diameter is given can be formed as
$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$
$\therefore[x-(-2)](x-3)+(y-3)[y-(-5)]=0$
$\Rightarrow (x+2)(x-3)+(y-3)(y+5)=0$
$\Rightarrow \left(x^2-x-6\right)+\left(y^2+2 y-15\right)=0$
$\Rightarrow x^2+y^2-x+2 y-21=0$

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Question 91 Mark

The equation of the circle which touches both the axes in the first quadrant and whose radius is $a$ is _________________ .

Answer

$x^2+y^2-2 a x-2 a y+a^2=0$, because
As the radius of circle is $a$ units and it touches both the axes in the first quadrant, its centre is $C(a, a)$.
Then equation of the circle is
$(x-a)^2+(y-a)^2=a^2$
$\Rightarrow x^2+y^2-2 a x-2 a y+a^2=0$

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