3 Marks Question

Question 13 Marks

Find the equation of the circle which passes through the point of intersection of the circles $x^2+y^2+2 x+3 y-7=0$ and $x^2+y^2-6 x+2 y-5=0$ and through the point $(2,-3)$.

Answer

Equation of any circle passing through the points of intersection of circles $S=0$ and $S^{\prime}=0$ is
$S+\lambda S^{\prime}=0$
Therefore, any circle passing through the points of intersection of the given circle is
$\left(x^2+y^2+2 x+3 y-7\right)+\lambda\left(x^2+y^2-6 x+2 y-5\right)=0\quad \ldots(i) $
The circle will pass through the point $(2,-3)$, if $(4+9+4-9-7)+\lambda(4+9-12-6-5)=0$
or, $\lambda=\frac{1}{10}$
Substituting the value of $\lambda$ in eq. (i), we get $\left(x^2+y^2+2 x+3 y-7\right)+\frac{1}{10}\left(x^2+y^2-6 x+2 y-5\right)=0$
or, $11 x^2+11 y^2+14 x+32 y-75=0$

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Question 23 Marks

Prove that the circles
$x^2+y^2+2 a x+c=0 \text { and } x^2+y^2+2 b y+c=0$
will touch each other, if
$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$

Answer

Centre of the first circle is $C_1(-a, 0)$ and its radius $=\sqrt{a^2-c}$
Centre of the second circle is $C_2(0,-b)$ and its radius $=\sqrt{b^2-c}$
The given circles will touch each other if the distance between their centres is equal to the sum or difference of their radii.
i.e., $C_1 C_2=r_1 \pm r_2$
$\begin{array}{rlrl}\Rightarrow & \sqrt{(-a-0)^2+(0+b)^2} =\sqrt{a^2-c} \pm \sqrt{b^2-c} \\ \Rightarrow & a^2+b^2 =\left(a^2-c\right)+\left(b^2-c\right) \pm 2 \sqrt{a^2-c} \cdot \sqrt{b^2-c} \\ \Rightarrow & 2 c= \pm 2 \sqrt{\left(a^2-c\right)\left(b^2-c\right)} \\ \Rightarrow & c^2= \left(a^2-c\right)\left(b^2-c\right) \\ & {[\text { on squaring both sides }]} \\ \Rightarrow & c^2= a^2 b^2-c\left(a^2+b^2\right)+c^2 \\ \Rightarrow & c\left(a^2+b^2\right)=a^2 b^2\\ \Rightarrow & \frac{1}{a^2}+\frac{1}{b^2} =\frac{1}{c} . \end{array}$
Hence Proved

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Question 33 Marks

Find the length of the chord intercepted by the circle $x^2+y^2-6 x+8 y-5=0$ on the line $2 x-y=5$.

Answer

The given line is
$2 x-y-5=0\quad \ldots(i) $
and the given circle is
$x^2+y^2-6 x+8 y-5=0\quad \ldots(ii) $
Its centre is $C(3,-4)$
and radius $=\sqrt{(3)^2+(4)^2-(-5)}=\sqrt{30}$. Let $d$ be perpendicular distance from $C(3,-4)$ on the line $(i)$, then
$\begin{aligned} d & =\frac{|2(3)-(-4)-5|}{\sqrt{(2)^2+(-1)^2}} \\ & =\frac{5}{\sqrt{5}}=\sqrt{5}\end{aligned}$
Here, $d<r$, so the line (i) and circle (ii) intersect in two points.
$\therefore$ The length of the chord intercepted
$\begin{array}{l}=2 \sqrt{r^2-d^2} \\ =2 \sqrt{(\sqrt{30})^2-(\sqrt{5})^2} \\ =2 \sqrt{30-5} \\ =2 \times 5=10 .\end{array}$

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Question 43 Marks

Find the equation of the circle which passes through the points $(0,3)$ and $(5,2)$ and whose centre lies on the $x$-axis.

Answer

Let the required equation of circle be
$x^2+y^2+2 g x+2 f y+c=0\quad \ldots(i) $
The centre of this circle is $(-g,-f)$. Since, given the centre of the circle lies on $X$-axis, the $Y$-coordinate of its centre will be zero, i.e., $-f=0$ or $f=0$
Therefore, eq. (i) can be written as
$x^2+y^2+2 g x+c=0\quad \ldots(ii) $
Since, the circle passes through the points $(0,3)$ and $(5,2)$, we get
$0+3^2+2 . g .0+c=0$
$
\begin{array}{l}
\text { or,} c=-9 \\
\text { and } 5^2+2^2+2 g(5)+c=0 \\
\text { or, } 25+4+10 g+c=0 \\
\text { or, } 10 g=-29-c \\
\text { or, } 10 g=-29-(-9) \\
\text { or, } 10 g=-20 \\
\Rightarrow g=-2
\end{array}
$
On substituting the values of $g$ and $c$ in eq. (ii), we get the required equation of circle
i.e., $ x^2+y^2-4 x-9=0$.

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Question 53 Marks

If $l x+m y= 1$ touches the circle $x^2+y^2=a^2$ then prove that the point $(l, m)$ lies on the circle $x^2+y^2$ $=a^{-2}$.

Answer

Since, the line $l x+m y-1=0$ touches the circle $x^2+$ $y^2=a^2$ (radius $=a$ ), so the perpendicular distance of the centre $(0,0)$ from the line is
$r=\frac{|0 . l+0 . m-1|}{\sqrt{l^2+m^2}}$
$\Rightarrow a=\frac{|-1|}{\sqrt{l^2+m^2}}$
$\Rightarrow \sqrt{l^2+m^2}=\frac{1}{a}$
$\Rightarrow l^2+m^2=a^{-2}$
Since, $(l, m)$ satisfies $x^2+y^2=a^{-2}$, so $(l, m)$ lies on the circle $x^2+y^2=a^{-2}$.

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Question 63 Marks

Find the cartesian equations of the curve $x=3 \cos t$, $y=-2+3 \sin t$. Do these equations represent a circle? If so, find centre and radius.

Answer

$\begin{array}{ll}\text {Given, } x=3 \cos t \text { and } y=-2+3 \sin t \\ \Rightarrow x=3 \cos t \text { and } y+2=3 \sin t\end{array}$
To eliminate the parameter $t$, on squaring and adding these equations, we get
$\begin{array}{ll} x^2+(y+2)^2=3^2\left(\cos ^2 t+\sin ^2 t\right) \\ \Rightarrow x^2+(y+2)^2=3^2\end{array}$
$\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]$
The equation $x^2+(y+2)^2=3^2$ represents a circle with centre $(0,-2)$ and radius $=3$ units.

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Question 73 Marks

Find the equations of circles which pass through the point $(3,6)$ and touch both the axes.

Answer

Let $a$ be the radius of the circle. Since, the circle passes through the point $P(3,6)$ and touches both the axes, it must lie in the first quadrant.
Therefore, the centre of the circle is $C(a, a)$.

Equation of the circle is
$(x-a)^2+(y-a)^2=a^2$
or, $x^2+y^2-2 a x-2 a y+a^2=0 \quad \ldots$
As the circle (i) passes through the point $(3,6)$, we get
$9+36-6 a-12 a+a^2=0$
$\Rightarrow a^2-18 a+45=0$
$\Rightarrow a^2-15 a-3 a+45=0$
$\Rightarrow (a-3)(a-15)=0$
$\Rightarrow a=3$ or $a=15$
If $a=3$, then the equation of circle is
$(x-3)^2+(y-3)^2=3^2$
$\Rightarrow x^2+y^2-6 x-6 y+9=0$
If $a=15$, then the equation of circle is
$(x-15)^2+(y-15)^2=15^2$
$\Rightarrow x^2+y^2-30 x-30 y+225=0$

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Question 83 Marks

Find the equation of the circle with radius 5 units whose centre lies on $X$-axis and passes through the point $(2,3)$.

Answer

As the centre of the circle lies on the $X$-axis, let its centre be $C(h, 0)$.
Since, the circle passes the point $A(2,3)$ and has radius 5.
$
\begin{array}{l}
\therefore \text { Radius,} C A=5 \\
\Rightarrow (2-h)^2+(3-0)^2=5^2 \\
\Rightarrow (2-h)^2=16 \\
\Rightarrow 2-h= \pm 4 \\
\Rightarrow h=-2 \text { or } h=6
\end{array}
$
$\therefore$ The centre of the circle is $(-2,0)$ or $(6,0)$
The equation of the circle is
$\begin{aligned}(x+2)^2+(y-0)^2 & =5^2 \\ \text { or } (x-6)^2+(y-0)^2 & =5^2\end{aligned}$
$\begin{array}{lr}\text { i.e., } x^2+y^2+4 x-21=0 \\ \text { or } x^2+y^2-12 x+11=0\end{array}$

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Question 93 Marks

Find the equation of the circle which passes through the point $(-2,-3)$ and has its centre on the negative direction of $X$-axis and is of radius 5 units.

Answer

As the centre of the circle lies on the negative direction of $X$-axis, let its centre be $C(h, 0), h<0$. Since, the circle passes through $A(-2,-3)$ and has radius 5 .
$\therefore C A=5$
$\begin{array}{lrl}\Rightarrow (h+2)^2+(0+3)^2 =5^2 \\ \Rightarrow (h+2)^2 =25-9\end{array}$
$\quad=16$
$\Rightarrow h+2= \pm 4$
$\Rightarrow h=2$ or $h=-6$
$\therefore$ The centre of the circle is $(-6,0)$ and its equation is
$\{x-(-6)\}^2+(y-0)^2=5^2$
$\begin{array}{l}\Rightarrow(x+6)^2+y^2=25 \\ \Rightarrow x^2+y^2+12 x+11=0 .\end{array}$

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Question 103 Marks

Find the equation of a circle having $(1,-2)$ as its centre and passing through intersection of the lines $3 x+y=14$ and $2 x+5 y=18$.

Answer

Given lines are :
$3 x+y=14$
and $2 x+5 y=18$
Solving eq (i) and (ii) simultaneously, we get
$x=4$ and $y=2$
$\therefore$ The point of intersection, say $P$, of the given lines is $(4,2)$.
Since, the centre of the circle is $C(1,-2)$ and it passes through the point $P(4,2)$ and its radius,
$C P=\sqrt{(4-1)^2+\{2-(-2)\}^2}$
$\begin{array}{l}=\sqrt{9+16} \\ =\sqrt{25} \\ =5 \text { unit }\end{array}$
$\therefore$ The required equation of circle is
$(x-1)^2+\{y-(-2)\}^2=5^2$
or,$\left(x^2-2 x+1\right)+\left(y^2+4 y+4\right)=25$
or, $x^2+y^2-2 x+4 y-20=0$.

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Question 113 Marks

Find the equation of the circle which is concentric with the circle $x^2+y^2-8 x-12 y+15=0$ and which passes through the point $(5,4)$.

Answer

Here, the centre of the given circle is $(4,6)$, so that the centre of the circle concentric with the given circle will also be $(4,6)$.
Therefore, the equation of circle concentric with the given circle will be of the form
$x^2+y^2-8 x-12 y+c=0\ldots(i) $
This circle will pass through the point $(5,4)$, if
$\begin{array}{rlrl} 5^2+4^2-8(5)-12(4)+c =0 \\ \text { or, } 25+16-40-48+c =0 \\ \text { or, } c =47\end{array}$
Hence, on substituting the value of $c$ in eq. (i), we get the required equation of circle.
$x^2+y^2-8 x-12 y+47=0$

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Question 123 Marks

Is the equation $2 x^2+2 y^2+3 y+10=0$ represents a circle? If so, find its centre and radius.

Answer

The given equation of circle is
$2 x^2+2 y^2+3 y+10=0$
On dividing by 2 , the equation can be written as
$x^2+y^2+\frac{3}{2} y+5=0$
One comparing the above equation with
$x^2+y^2+2 g x+2 f y+c=0$, we get
$g=0, f=\frac{3}{4}$ and $c=5$
Now, $g^2+f^2-c$
$
\begin{array}{l}
=0+\left(\frac{3}{4}\right)^2-5 \\
=\frac{9}{16}-5 \\
=-\frac{71}{16}<0
\end{array}
$
Hence, the given equation represents the empty set i.e., the radius of the circle is imaginary and it is not possible to draw the circle.

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Question 133 Marks

Find the centre and the radius of the circle $2 x^2+$ $2 y^2+10 x-6 y-1=0$

Answer

Given equation of circle is
$2 x^2+2 y^2+10 x-6 y-1=0$
One dividing by 2 , the equation of circle can be written as
$x^2+y^2+5 x-3 y-\frac{1}{2}=0$
On comparing above equation with
$
\begin{array}{l}
x^2+y^2+2 g x+2 f y+c=0, \text { we get } \\
g=\frac{1}{2}(\text { coefficient of } x)=\frac{1}{2}(5)=\frac{5}{2} \\
f=\frac{1}{2}(\text { coefficient of } y)=\frac{1}{2}(-3)=-\frac{3}{2}
\end{array}
$
and $c =$ constant term $=-\frac{1}{2}$
Hence, the centre of the circle is $(-g,-f)$ i.e., $\left(-\frac{5}{2}, \frac{3}{2}\right)$
and radius of the circle
$\begin{array}{l}=\sqrt{g^2+f^2-c} \\ =\sqrt{\left(\frac{5}{2}\right)^2+\left(-\frac{3}{2}\right)^2-\left(-\frac{1}{2}\right)} \\ =\sqrt{\frac{25}{4}+\frac{9}{4}+\frac{1}{2}} \\ =\sqrt{\frac{36}{4}}=3 .\end{array}$

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Question 143 Marks

Find the equation of the circle passing through $(0,0)$ and which makes intercepts $a$ and $b$ on the coordinate axes.

Answer

Since, the circle passes through $O(0,0)$ and cuts off intercepts $a$ and $b$ on the co-ordinate axes, it passes through $A(0, a)$ and $B(0, b)$.
As the axes are perpendicular to each other, Therefore $\angle A O B=90^{\circ}$ and hence $A B$ becomes a diameter of this circle.

$\therefore$ The equation of the circle is given by
$(x-a)(x-0)+(y-0)(y-b)=0$
or, $x^2+y^2-a x-b y=0$

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