Exercise 10.1 (Revised) - Chapter 11 - Conic Sections - Ncert Solutions class 11 - Maths

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Chapter 10: Conic Sections - NCERT Solutions for Class 11 Maths

In each of the following Exercises 1 to 5, find the equation of the circle with:
Ex 10.1 Question 1.

Centre $(0,2)$ and radius 2 .

Answer.

Given: $h=0, k=2$ and $\gamma=2$
Equation of the circle;
$
\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& \therefore(x-0)^2+(y-2)^2=(2)^2 \\
& \Rightarrow x^2+y^2+4-4 y=4 \\
& \Rightarrow x^2+y^2-4 y=0
\end{aligned}
$
Ex 10.1 Question 2.

Centre $(-2,3)$ and radius 4 .

Answer.

Given: $h=-2, k=3$ and $\mu=4$
Equation of the circle ;
$
\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& \therefore(x+2)^2+(y-3)^2=(4)^2
\end{aligned}
$

$\begin{aligned}
& \Rightarrow x^2+4+4 x+y^2+9-6 y=16 \\
& \Rightarrow x^2+y^2+4 x-6 y-3=0
\end{aligned}$

Ex 10.1 Question 3.

Centre $\left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $\frac{1}{12}$

Answer.

Given: $h=\frac{1}{2}, k=\frac{1}{4}$ and $\gamma=\frac{1}{12}$
Equation of the circle;
$
\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& \therefore\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{4}\right)^2=\left(\frac{1}{12}\right)^2 \\
& \Rightarrow x^2+\frac{1}{4}-x+y^2+\frac{1}{16}-\frac{1}{2} y=\frac{1}{144} \\
& \Rightarrow 144 x^2+36-144 x+144 y^2+9-72 y=1 \\
& \Rightarrow 144 x^2+144 y^2-144 x-72 y+44=0 \\
& \Rightarrow 4\left(36 x^2+36 y^2-36 x-18 y+11\right)=0 \\
& \Rightarrow 36 x^2+36 y^2-36 x-18 y+11=0
\end{aligned}
$

Ex 10.1 Question 4.

Centre (1,1) and radius $\sqrt{2}$.

Answer.

Given: $h=1, k=1$ and $r=\sqrt{2}$
Equation of the circle;
$
\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& \therefore(x-1)^2+(y-1)^2=(\sqrt{2})^2 \\
& \Rightarrow x^2+1-2 x+y^2+1-2 y=2
\end{aligned}
$$

Ex 10.1 Question 5.

Centre $(-a,-b)$ and radius $\sqrt{a^2-b^2}$.

Answer.

Given: $h=-a, k=-b$ and $r=\sqrt{a^2-b^2}$
Equation of the circle ;
$
\begin{aligned}
& (x-h)^2+(y-k)^2=r^2 \\
& \therefore(x+a)^2+(y+b)^2=\left(\sqrt{a^2-b^2}\right)^2 \\
& \Rightarrow x^2+a^2+2 a x+y^2+b^2+2 b y=a^2-b^2 \\
& \Rightarrow x^2+y^2+2 a x+2 b y+2 b^2=0
\end{aligned}
$

In each of the following Exercises 6 to 9, find the centre and radius of the circles.
Ex 10.1 Question 6.

$(x+5)^2+(y-3)^2=36$

Answer.

Given: Equation of the circle;
$
\begin{aligned}
& (x+5)^2+(y-3)^2=36 \\
& \Rightarrow(x+5)^2+(y-3)^2=(6)^2
\end{aligned}
$

On comparing eq. (i) with $(x-h)^2+(y-k)^2=r^2$ $h=-5, k=3$ and $\gamma=6$
Ex 10.1 Question 7.

$x^2+y^2-4 x-8 y-45=0$

Answer.

Given: Equation of the circle: $x^2+y^2-4 x-8 y-45=0$

$
\begin{aligned}
& \Rightarrow\left(x^2-4 x\right)+\left(y^2-8 y\right)=45 \\
& \Rightarrow\left(x^2-4 x+2^2\right)+\left(y^2-8 y+4^2\right)=45+2^2+4^2 \\
& =>(x-2)^2+(y-4)^2=(\sqrt{65})^2 \ldots \ldots .(i)
\end{aligned}
$

On comparing eq. (i) with $(x-h)^2+(y-k)^2=r^2$
$
h=2, k=4 \text { and } r=\sqrt{65}
$

Ex 10.1 Question 8.

$x^2+y^2-8 x-10 y-12=0$

Answer.

Given: Equation of the circle;
$
\begin{aligned}
& =>\left(x^2-8 x\right)+\left(y^2-10 y\right)=12 \\
& =>\left(x^2-8 x+4^2\right)+\left(y^2-10 y+5^2\right)=12+4^2+5^2 \\
& \Rightarrow(x-4)^2+(y-5)^2=(\sqrt{53})^2 \\
&
\end{aligned}
$

On comparing eq. (i) with $(x-h)^2+(y-k)^2=r^2$
We get, $\mathrm{h}=4, \mathrm{k}=5$ and $r=\sqrt{53}$
Ex 10.1 Question 9.

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

Answer.

Given: Equation of the circle: $2 x^2+2 y^2-x=0$
$\Rightarrow x^2+y^2-\frac{x}{2}=0 \quad$ (divide the equation by 2 )
$\Rightarrow\left(x^2-\frac{x}{2}\right)+y^2=0$

$\Rightarrow\left[x^2-\frac{x}{2}+\left(\frac{1}{4}\right)^2\right]+y^2=0+\left(\frac{1}{4}\right)^2$

$
\Rightarrow\left(x-\frac{1}{4}\right)^2+(y-0)^2=\left(\frac{1}{4}\right)^2 \ldots \ldots . . \text { (i) }
$

On comparing eq. (i) with $(x-h)^2+(y-k)^2=r^2$
$
h=\frac{1}{4}, k=0 \text { and } r=\frac{1}{4}
$

Ex 10.1 Question 10.

Find the equation of the circle passing through the points $(4,1)$ and $(6,5)$ and whose centre lies on the line $4 x+y=16$.

Answer.

The equation of the circle is $(x-h)^2+(y-k)^2=r^2$ $\qquad$
$\because$ Circle passes through point $(4,1)$
$
\begin{aligned}
& \therefore(4-h)^2+(1-k)^2=r^2 \\
& \Rightarrow 16+h^2-8 h+1+k^2-2 k=r^2 \\
& \Rightarrow h^2+k^2-8 h-2 k+17=r^2 \ldots .
\end{aligned}
$

Again Circle passes through point $(6,5)$
$
\begin{aligned}
& \therefore(6-h)^2+(5-k)^2=r^2 \\
& \Rightarrow 36+h^2-12 h+25+k^2-10 k=r^2 \\
& \Rightarrow h^2+k^2-12 h-10 k+61=r^2 \ldots \ldots .
\end{aligned}
$

From eq. (ii) and (iii), we have
$
h^2+k^2-8 h-2 k+17=h^2+k^2-12 h-10 k+61
$

$
\begin{aligned}
& \Rightarrow 4 h+8 k=44 \\
& \Rightarrow h+2 k=11 \ldots
\end{aligned}
$
Since the centre $(h, k)$ of the circle lies on the line $4 x+y=16$

$
\therefore 4 h+k=16
$

On solving eq. (iv) and (v), we have $h=3, k=4$
Putting the values of $h$ and $k$ in eq. (ii), we have
$
\begin{aligned}
& 3^2+4^2-8 \times 3-2 \times 4+17=r^2 \\
& \Rightarrow r^2=9+16-24-8+17=10
\end{aligned}
$

Therefore, the equation of the required circle is
$
\begin{aligned}
& (x-3)^2+(y-4)^2=10 \\
& \Rightarrow x^2+9-6 x+y^2+16-8 y=10 \\
& \Rightarrow x^2+y^2-6 x-8 y+15=0
\end{aligned}
$
Ex 10.1 Question 11.

Find the equation of the circle passing through the points $(2,3)$ and $(-1,1)$ and whose centre lies on the line $x-3 y-11=0$.

Answer.

The equation of the circle is $(x-h)^2+(y-k)^2=r^2$
$\because$ Circle passes through point $(2,3)$
$
\therefore(2-h)^2+(3-k)^2=r^2
$

$\begin{aligned}
& \Rightarrow 1+h^2+2 h+1+k^2-2 k=r^2 \\
& \Rightarrow h^2+k^2+2 h-2 k+2=r^2 \ldots
\end{aligned}$

From eq. (ii) and (iii), we have
$
\begin{aligned}
& =>h^2+k^2-4 h-6 k+13=h^2+k^2+2 h-2 k+2 \\
& \Rightarrow-6 h-4 k=-11 \\
& \Rightarrow 6 h+4 k=11 \text {........(iv) }
\end{aligned}
$

Since the centre $(h, k)$ of the circle lies on the line $x-3 y-11=0$
$
\therefore h-3 k=11 \text {. }
$

On solving eq. (iv) and (v), we have $h=\frac{7}{2}, k=\frac{-5}{2}$
Putting the values of $h$ and $k$ in eq. (ii), we have
$
\begin{aligned}
& \left(\frac{7}{2}\right)^2+\left(\frac{-5}{2}\right)^2-\frac{4 \times 7}{2}-6 \times \frac{-5}{2}+13=r^2 \\
& \Rightarrow \frac{49}{4}+\frac{25}{4}-14+15+13=r^2 \\
& \Rightarrow r^2=\frac{65}{2}
\end{aligned}
$

Therefore, the equation of the required circle is

$\begin{aligned}
& \left(x-\frac{7}{2}\right)^2+\left(y+\frac{5}{2}\right)^2=\frac{65}{2} \\
& \Rightarrow x^2+\frac{49}{4}-7 x+y^2+\frac{25}{4}+5 y=\frac{65}{2} \\
& \Rightarrow 4 x^2+49-28 x+4 y^2+25+20 y=130 \\
& \Rightarrow 4 x^2+4 y^2-28 x+20 y-56=0 \\
& \Rightarrow 4\left(x^2+y^2-7 x+5 y-14\right)=0
\end{aligned}$

$
\Rightarrow x^2+y^2-7 x+5 y-14=0
$
Ex 10.1 Question 12.

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

Answer.

Since the centre of circle lies on $x$-axis, therefore the coordinates of centre is $(h, 0)$.
Now the circle passes through the point $(2,3)$. According to the question,
$
\begin{aligned}
& \sqrt{(h-2)^2+(0-3)^2}=5 \\
& \Rightarrow \sqrt{h^2+4-4 h+9}=5 \\
& \Rightarrow h^2+4-4 h+9=25 \\
& \Rightarrow h^2-4 h-12=0 \\
& \Rightarrow(h-6)(h+2)=0 \\
& \Rightarrow h=6 \text { or } h=-2
\end{aligned}
$

Taking $h=6$, Equation of the circle is $(x-6)^2+(y-0)^2=(5)^2$

$
\begin{aligned}
& \Rightarrow x^2+36-12 x+y^2=25 \\
& \Rightarrow x^2+y^2-12 x+11=0
\end{aligned}
$

Taking $h=-2$, Equation of the circle is $(x+2)^2+(y-0)^2=(5)^2$
$
\begin{aligned}
& \Rightarrow x^2+4+4 x+y^2=25 \\
& \Rightarrow x^2+y^2+4 x-21=0
\end{aligned}
$
Ex 10.1 Question 13.

Find the equation of the circle passing through (0, 0) and making intercept $a$ and $b$ on the coordinate axes.

Answer

$\text {The circle makes intercepts } a \text { with } x \text {-axis and } b \text { with } y \text {-axis. }$

$\therefore \mathrm{OA}=a$ and $\mathrm{OB}=b$
$\therefore$ Coordinates of A and B are $(a, 0)$ and $(0, b)$ respectively.
Now the circle passes through the points $\mathrm{O}(0,0), \mathrm{A}(a, 0)$ and $\mathrm{B}(0, b)$.
Putting these coordinates of three points in the equation of the circle,
$
x^2+y^2+2 g x+2 f y+c=0
$

Circle passing through $(0,0)$
$
\Rightarrow c=0
$
the circle also passes through $(\mathrm{a}, 0$ ) and $(0, \mathrm{~b})$
$
\begin{aligned}
& a^2+2 g a=0 \\
& \Rightarrow a(a+2 g)=0 \\
& \Rightarrow g=\frac{-1}{2} a
\end{aligned}
$

And $b^2+2 f b=0$
$
\Rightarrow b(b+2 f)=0
$

$
\Rightarrow f=\frac{-1}{2} b
$

Putting the values of $g, f$ and $c$ in eq. (i), we have
$
\begin{aligned}
& x^2+y^2+2 \times \frac{-1}{2} a x+2 \times \frac{-1}{2} b y+0=0 \\
& \Rightarrow x^2+y^2-a x-b y=0
\end{aligned}
$
Ex 10.1 Question 14.

Find the equation of the circle with centre $(2,2)$ and passes through the point $(4,5)$.

Answer.

The equation of the circle is $(x-h)^2+(y-k)^2=r^2$
Since the circle passes through point $(4,5)$ and coordinates of centre are $(2,2)$.
$\therefore$ Radius of circle $=\sqrt{(4-2)^2+(5-2)^2}=\sqrt{4+9}=\sqrt{13}$
Therefore, the equation of the required circle is
$
\begin{aligned}
& (x-2)^2+(y-2)^2=(\sqrt{13})^2 \\
& \Rightarrow x^2+4-4 x+y^2+4-4 y=13 \\
& \Rightarrow x^2+y^2-4 x-4 y-5=0
\end{aligned}
$

Ex 10.1 Question 15.

Does the point $(-2.5,3.5)$ lie inside, outside or on the circle $x^2+y^2=25$ ?

Answer.

Given: Equation of the circle $x^2+y^2=25$
$
\Rightarrow(x-0)^2+(y-0)^2=(5)^2
$

On comparing with $(x-h)^2+(y-k)^2=r^2$, we have $h=0, k=0$ and $r=5$
Now distance of the point $(-2.5,3.5)$ from the centre $(0,0)$
$
=\sqrt{(0+2.5)^2+(0-3.5)^2}=\sqrt{6.25+12.25}=\sqrt{18.5}=4.3<
$

Therefore, the point $(-2.5,3.5)$ lies inside the circle.