Exercise 5.1 - Chapter 5 - Two Dimensional Analytical Geometry – II - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Two Dimensional Analytical Geometry - II
$\operatorname{Ex} 5.1$
Question 1.
Obtain the equation of the circles with radius $5 \mathrm{~cm}$ and touching $\mathrm{x}$-axis at the origin in general form.

Solution:
Given radius $=5 \mathrm{~cm}$ and the circle is touching $\mathrm{x}$ axis
So centre will be $(0, \pm 5)$ and radius $=5$
The equation of the circle with centre $(0, \pm 5)$ and radius 5 units is $(\mathrm{x}-0)^2+(\mathrm{y} \pm 5)^2=5^2$
(i.e) $x^2+y^2 \pm 10 y+25-25=0$
(i.e) $x^2+y^2 \pm 10 y=0$
Question 2.
Find the equation of the circle with centre $(2,-1)$ and passing through the point $(3,6)$ in standard form.
Solution:
Centre $=\mathrm{C}=(2,-1) ;$ Passing through $=\mathrm{A}=(3,6)$
So radius $=\mathrm{CA}=\sqrt{(2-3)^2+(-1-6)^2}=\sqrt{1+49}=5 \sqrt{50}$
Now centre $=(2,-1)$ and radius $=\sqrt{50}$
So equation of the circle is
(i.e) $(x-2)^2+(y+1)^2=\sqrt{50}^2 \Rightarrow(x-2)^2+(y+1)^2=50$
Question 3.
Find the equation of circles that touch both the axes and pass through $(-4,-2)$ in general form.
Solution:
Since the circle touches both the axes, its centre will be $(r, r)$ and radius wil be $r$.
Here centre $=\mathrm{C}=(\mathrm{r}, \mathrm{r})$ and point on the circle is $\mathrm{A}=(-4,-2)$
$
\mathrm{CA}=\mathrm{r} \Rightarrow \mathrm{CA}^2=\mathrm{r}^2
$

$
\begin{aligned}
& \text { (i.e) }(r+4)^2+(r+2)^2=r^2 \\
& \Rightarrow r^2+8 r+16+r^2+4 r+4-r^2=0 \\
& \text { (i.e) } r^2+12 r+20=0 \\
& (r+2)(r+10)=0 \\
& \Rightarrow r=-2 \text { or }-10
\end{aligned}
$
When $r=-2$, the equation of the circle will be $(x+2)^2+(y+2)^2=2^2$
$
\text { (i.e) } x^2+y^2+4 x+4 y+4=0
$
When $r=-10$, the equation of the circle will be $(x+10)^2+(y+10)^2=10^2$
$
\text { (i.e) } x^2+y^2+20 x+20 y+100=0
$

Question 4.
Find the equation of the circle with centre $(2,3)$ and passing through the intersection of the lines $3 \mathrm{x}-2 \mathrm{y}-1=$ 0 and $4 x+y-27=0$
Solution:
To find the point of intersection of the two lines we have to solve the two equations.
Now solving them: $3 x-2 y=1$ $4 \mathrm{x}+\mathrm{y}=27 \ldots \ldots . .(2)$
(2) $\times 2 \Rightarrow 8 \mathrm{x}+2 \mathrm{y}=54$
(1) $\Rightarrow 3 \mathrm{x}-2 \mathrm{y}=1$
(3) $+(1) \Rightarrow 11 \mathrm{x}=55 \Rightarrow \mathrm{x}=\frac{55}{11}=5$
Substituting $x=5$ in (2) we get
$20+\mathrm{y}=27 \Rightarrow \mathrm{y}=27-20=7$
$\therefore$ The point $=\mathrm{A}=(5,7)$
Given centre $=\mathrm{C}=(2,3)$
$\therefore$ radius $=\sqrt{(5-2)^2+(7-3)^2}=\sqrt{9+16}=5=5$
Now centre $=(2,3)$ and radius $=5$
So equation of the circle is
$(\mathrm{x}-2)^2+(\mathrm{y}-3)^2=5^2$
(i.e) $x^2+y^2-4 x-6 y-12=0$

Question 5.
Obtain the equation of the circle for which $(3,4)$ and $(2,-7)$ are the ends of a diameter.
Solution:
The equation of a circle with $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ as end points of a diameter is $\left(\mathrm{x}-\mathrm{x}_1\right)\left(\mathrm{x}-\mathrm{x}_2\right)+\left(\mathrm{y}-\mathrm{y}_1\right)\left(\mathrm{y}-\mathrm{y}_2\right)=0$
Here the end points of a diameter are $(3,4)$ and $(2,-7)$
So equation of the circle is $(x-3)(x-2)+(y-4)(y+7)=$.
$
x^2+y^2-5 x+37-22=0
$

Question 6.
Find the equation of the circle through the points $(1,0),(-1,0)$, and $(0,1)$.
Solution:
Let the required circle be
$
x^2+y^2+2 g x+2 f y+c=0
$
The circle passes through $(1,0),(-1,0)$ and $(0,1)$
$
\begin{aligned}
& (1,0) \Rightarrow 1+0+2 \mathrm{~g}(1)+2 \mathrm{f}(0)+\mathrm{c}=0 \\
& 2 \mathrm{~g}+\mathrm{c}=-1 \ldots \ldots \ldots \ldots(1) \\
& (-1,0) \Rightarrow 1+0+2 \mathrm{~g}(-1)+2 \mathrm{f}(0)+\mathrm{c}=0
\end{aligned}
$
$
\begin{aligned}
& -2 \mathrm{~g}+\mathrm{c}=-1 \ldots \ldots \ldots(2) \\
& (0,1) \Rightarrow 0+1+2 \mathrm{~g}(0)+2 \mathrm{f}(1)+\mathrm{c}=0 \\
& 2 \mathrm{f}+\mathrm{c}=-1 \ldots \ldots \ldots(3) \\
& \text { Now solving (1), (2) and (3) } \\
& 2 \mathrm{~g}+\mathrm{c}=-1 \ldots \ldots \ldots \ldots(1) \\
& -2 \mathrm{~g}+\mathrm{c}=-1 \ldots \ldots \ldots \ldots(2) \\
& (1)+(2) \Rightarrow 2 \mathrm{c}=-2 \Rightarrow \mathrm{c}=-1
\end{aligned}
$
Substituting $c=-1$ in (1) we get
$
\begin{aligned}
& 2 \mathrm{~g}-1=-1 \\
& 2 \mathrm{~g}=-1+1=0 \Rightarrow \mathrm{g}=0
\end{aligned}
$
Substituting c $=-1$ in (3) we get
$
2 \mathrm{f}-1=-1 \Rightarrow 2 \mathrm{f}=-1+1=0 \Rightarrow \mathrm{f}=0
$
So we get $g=0, f=0$ and $c=-1$
So the required circle will be
$
\begin{aligned}
& x^2+y^2+2(0) x+2(0) y-1=0 \\
& \text { (i.e) } x^2+y^2-1=0 \Rightarrow x^2+y^2=1
\end{aligned}
$

Question 7.
A circle of area $9 n$ square units has two of its diameters along the lines $x+y=5$ and $x-y=1$. Find the equation of the circle.
Solution:
Area of the circle $=9 \pi$
(i.e) $\pi r^2=9 \pi$ $\Rightarrow \mathrm{r}^2=9 \Rightarrow \mathrm{r}=3$
(i.e) radius of the circle $=r=3$
The two diameters are $x+y=5$ and $x-y=1$
The point of intersection of the diameter is the centre of the circle $=\mathrm{C}$
To find $\mathrm{C}$ : Solving $\mathrm{x}+\mathrm{y}=5$ $\mathrm{x}-\mathrm{y}=1$
(1) $+(2) \Rightarrow 2 \mathrm{x}=6 \Rightarrow \mathrm{x}=3$
Substituting $x=3$ in (1) we get
$3+y=5 \Rightarrow y=5-3=2$
$\therefore$ Centre $=(3,2)$ and radius $=3$
So equation of the circle is $(x-3)^2+(y-2)^2=3^2$
(i.e) $x^2+y^2-6 x-4 y+4=0$

Question 8.
If $\mathrm{y}=2 \sqrt{2} x+\mathrm{c}$ is a tangent to the circle $\mathrm{x}^2+\mathrm{y}^2=16$, find the value of $\mathrm{c}$.
Solution:
The condition for the line $y=m x+c$ to be a tangent to the circle $x^2+y^2=a^2$ is $c^2=a^2\left(1+m^2\right)$
Here $\mathrm{x}^2+\mathrm{y}^2=16 \Rightarrow \mathrm{a}^2=16$
$\mathrm{y}=2 \sqrt{2} x+\mathrm{c} \Rightarrow \mathrm{m}=2 \sqrt{2}$ and $\mathrm{c}=\mathrm{c}$
The condition is $\mathrm{c}^2=\mathrm{a}^2\left(1+\mathrm{m}^2\right)$
(i.e) $c^2=16(1+8)=144$
$\Rightarrow \mathrm{c}=\pm 12$

Question 9.
Find the equation of the tangent and normal to the circle $x^2+y^2-6 x+6 y-8=0$ at $(2,2)$.
Solution:
The equation of the tangent to the circle $x^2+y^2+2 g x+2 f y+c=0$ at $\left(x_1, y_1\right)$ is $x_1+y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=9$
So the equation of the tangent to the circle $\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+6 \mathrm{y}-8=0$ at $\left(x_1, y_1\right)$ is
$
\begin{aligned}
& \mathrm{xx}_1+\mathrm{yy}_1-\frac{6\left(x+x_1\right)}{2}+\frac{6\left(y+y_1\right)}{2}-8=0 \\
& \text { (i.e) } \mathrm{xx}_1+\mathrm{yy}_1-3\left(\mathrm{x}+\mathrm{x}_1\right)+3\left(\mathrm{y}+\mathrm{y}_1\right)-8=0 \\
& \text { Here }\left(\mathrm{x}_1, \mathrm{y}_1\right)=(2,2)
\end{aligned}
$
So equation of the tangent is
$\mathrm{x}(2)+\mathrm{y}(2)-3(\mathrm{x}+2)+3(\mathrm{y}+2)-8=0$
(i.e) $2 \mathrm{x}+2 \mathrm{y}-3 \mathrm{x}-6+3 \mathrm{y}+6-8=0$
(i.e) $-x+5 y-8=0$ or $x-5 y+8=0$
Normal is a line $\perp r$ to the tangent
So equation of normal circle be of the form $5 x+y+k=0$
The normal is drawn at $(2,2)$
$
\Rightarrow 10+2+\mathrm{k}=0 \Rightarrow \mathrm{k}=-12
$
So equation of normal is $5 x+y-12=0$

Question 10.
Determine whether the points $(-2,1),(0,0)$ and $(-4,-3)$ lie outside, on or inside the circle $x^2+y^2-5 x+2 y-$ $5=0$.
Solution:
To find the position of a point with regard to a given circle, substitute the point in the equation of the circle if we get a positive value, the point lies outside the circle.
If we get a -ve value the point lies inside the circle and if we get $O$ then the point lies on the circumference of the circle.
The given circle is $x^2+y^2-5 x+2 y-5=0$
Substituting the point $(-2,1)$ in (1) we get
$4+1-5(-2)+2(1)-5=5+10+2-5=12$
$\Rightarrow(-2,1)$ lies outside the circle
Substituting the point $(0,0)$ in (1) we get
$-5<0 \Rightarrow(0,0)$ lies inside the circle Substituting the point $(-4,-3)$ in (1) we get $16+9+20-6-5=34>0$
$\Rightarrow(-4,-3)$ lies outside the circle

Question 11.

Find centre and radius of the following circles.
(i) $x^2+(y+2)^2=0$
(ii) $x^2+y^2+6 x-4 y+4=0$
(iii) $x^2+y^2-x+2 y-3=0$
(iv) $2 x^2+2 y^2-6 x+4 y+2=0$
Solution:
(i) $x^2+(y+2)^2=0$
(i.e) $x^2+y^2+4 y+4=0$
Comparing this equation with the general form $x^2+y^2+2 g x+2 f y+c=0$
we get $2 \mathrm{~g}=0 \Rightarrow \mathrm{g}=0$
$2 \mathrm{f}=4 \Rightarrow \mathrm{f}=2$ and $\mathrm{c}=4$
Now centre $=(-\mathrm{g},-\mathrm{f})=(0,-2)$

$
\begin{aligned}
& \text { Radius }=\mathrm{r}=\sqrt{g^2+f^2-c}=\sqrt{0+4-4} \\
& \therefore \text { Centre }=(0,-2) \text { and radius }=0
\end{aligned}
$
(ii) $x^2+y^2+6 x-4 y+4=0$
Comparing with the general form we get
$
\begin{aligned}
& 2 \mathrm{~g}=6,2 \mathrm{f}=-4 \\
& \Rightarrow \mathrm{g}=3, /=-2 \text { and } \mathrm{c}=4 \\
& \text { Centre }=(-\mathrm{g},-\mathrm{f})=(-3,2) \\
& \text { Radius }=\sqrt{g^2+f^2-c}=\sqrt{9+4-4}=3 \\
& \therefore \text { Centre }=(-3,2) \text { and radius }=3
\end{aligned}
$

(iii) $x^2+y^2-x+2 y-3=0$
Comparing with the general form of the circle we get
$
\begin{array}{l|l|l|l}
2 g=-1 & 2 f=2 & \\
g=-\frac{1}{2} & f=1 & c=-3
\end{array}
$
$
\begin{aligned}
\text { So centre }=(-g,-f) & =\left(\frac{1}{2},-1\right) \\
\text { Radius } & =\sqrt{g^2+f^2-c}=\sqrt{\frac{1}{4}+1+3}=\sqrt{\frac{17}{2}} \\
\therefore \quad \text { Centre } & =\left(\frac{1}{2},-1\right) \text { and radius }=\sqrt{\frac{17}{2}}
\end{aligned}
$

(iv) $2 x^2+2 y^2-6 x+4 y+2=0$
$(\div$ by 2$) \Rightarrow x^2+y^2-3 x+2 y+1=0$
Comparing this equation with the general form of the circle we get
$2 \mathrm{~g}=-3,2 \mathrm{f}=2$
$\mathrm{g}=-\frac{3}{2}, \mathrm{~g}=1$ and $\mathrm{c}=1$
So centre $=(-\mathrm{g},-\mathrm{f})=\left(\frac{3}{2},-1\right)$
and radius $=\sqrt{g^2+f^2-c}=\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}$
$\therefore$ Centre $=\left(\frac{3}{2},-1\right)$ and radius $=\frac{3}{2}$
Question 12.

If the equation $3 x$
${ }^2+(3-p) x y+q y^2-2 p x=8 p q$ represents a circle, find $p$ and $q$. Also determine the centre and radius of the circle.
Solution:
For a circle co-eff of $x^2=$ co-efif of $y^2$
$
\Rightarrow 3=\mathrm{q}
$
co-eff of $x y=0$
$
\Rightarrow 3-\mathrm{p}=0 \Rightarrow \mathrm{p}=3
$
So $\mathrm{p}=\mathrm{q}=3$
So the equation of the circle becomes $3 x^2+3 y^2-6 x-72=0$
$
(\div \text { by } 3) \Rightarrow x^2+y^2-2 x-24=0
$

Comparing this equation with the general form of the circle we get
$
\begin{aligned}
& 2 \mathrm{~g}=-2,2 \mathrm{f}=0 \\
& \mathrm{~g}=-1, \mathrm{f}=0 \text { and } c=-24
\end{aligned}
$
So centre $=(-\mathrm{g},-\mathrm{f})=(1,0)$ and radius $=\sqrt{g^2+f^2-c}=5$
$\therefore$ Centre $=(1,0)$ and radius $=5$