Exercise 5.1-Additional Problems - Chapter 5 Two Dimensional Analytical Geometry–II 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Problems
Question 1.
Find the equation of the circle whose centre is $(2,-3)$ and passing through the intersection of the line $3 x-2 y$ $=1$ and $4 x+y=27$.
Solution:
Solving $3 x-2 y=1$ and $4 x+y=27$
Simultaneously, we get $x=5$ and $y=7$
$\therefore$ The point of intersection of the lines is $(5,7)$

Now we have to find the equation of a circle whose centre is $(2,-3)$ and which passes through $(5,7)$
$
\begin{aligned}
\text { Radius } & =\sqrt{(5-2)^2+(7+3)^2} \\
& =\sqrt{9+100}=\sqrt{109}
\end{aligned}
$
$\therefore$ Required equation of the circle is
$
\begin{aligned}
& (x-2)^2+(y+3)^2=(\sqrt{109})^2 \\
& \Rightarrow x^2+y^2-4 x+6 y-96=0
\end{aligned}
$
Question 2.
Find the equation of a circle of radius 5 whose centre lies on $\mathrm{x}$-axis and which passes through the point $(2,3)$.
Solution:
Let the centre of the circle be $(h, k)$. Since the centre lies on $\mathrm{x}$-axis, we have $k=0$
Therefore, centre $=(\mathrm{h}, 0)$
Also the circle has radius 5 units and passes through the point $(2,3)$.
$
\begin{aligned}
& \therefore \quad \sqrt{(2-h)^2+(3-0)^2}=5 \Rightarrow h^2-4 h+4+9=25 \\
& \Rightarrow \quad h^2-4 h-12=(h-6)(h+2)=0 \\
& \therefore h=6,-2
\end{aligned}
$
Hence the centre is $(6,0)$ or $(-2,0)$

$\therefore$ The equation of the circle is
$
(x-6)^2+(y-0)^2=5^2 \Rightarrow x^2+y^2-12 x+11=0(\text { Or })(x+2)^2+(y-0)^2=5^2 \Rightarrow x^2+y^2+4 x-21=0
$
Question 3.
Find the centre and radius of the following circles:
$x^2+y^2-2 x+4 y-4=0$ and $2 x^2+2 y^2+16 x-28 y+32=0$. Also find the ratio of their diameters.
Solution:
Comparing the equation $x^2+y^2-2 x+4 y-4=0$ with $x^2+y^2+2 g x+2 f y+c=0$, we get $2 \mathrm{~g}=-2 \Rightarrow \mathrm{g}=-1,2 \mathrm{f}=4 \Rightarrow \mathrm{f}=2, \mathrm{c}=-4$
Centre $=(-g,-f)=(1,-2)$ and radius $r_1=\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3$ units
Diameter $=2 r_1=2 \times 3=6$ units
Again $2 x^2+2 y^2+16 x-28 y+32=0$
$(\div$ by 2$) \Rightarrow x^2+y^2+8 x-14 y+16=0$
Comparing with $x^2+y^2+2 g x+2 f y+c=0$
$2 \mathrm{~g}=8 \Rightarrow \mathrm{g}=4,2 \mathrm{f}=-14 \Rightarrow \mathrm{f}=-7, \mathrm{c}=16$
radius $r_2=\sqrt{g^2+f^2-c}=\sqrt{16+49-16}=\sqrt{49}=7$ units
Diameter $=2 r_2=2 \times 7=14$ units
Ratio of their diameters $=6: 14=3: 7$.
Question 4.
Find the equation of the circle whose radius is 4 and which is concentric with the circle $x^2+y^2+2 x-6 y=0$

Solution:
$
x^2+y^2+2 x-6 y=0
$
Here $2 \mathrm{~g}=2 \Rightarrow \mathrm{g}=1,2 \mathrm{f}=-6 \Rightarrow \mathrm{f}=-3$
Centre of the circle $=(-g,-f)=(-1,3)$
Since the required circle is concentric with (1), its centre is also $(-1,3)$.
$\therefore$ The equation of the circle whose centre is $(-1,3)$ and radius 4 is
$(x+1)^2+(y-3)^2=4^2$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-6 \mathrm{y}-6=0$
Question 5.
Show that the four points $(1,0),(2,-7),(8,1)$ and $(9,6)$ are concyclic.
Solution:

Let the equation of the circle be Since it passes through $(1,0) \Rightarrow x^2+y^2+2 g x+2 f y+c=0$
Since it passes through $(1,0)$
$
\therefore \quad 1+0+2 g+0+c=0 \quad \Rightarrow 2 g+c+1=0
$
Since it passes through $(2,-7)$
$
\therefore 4+49+4 g-14 f+c=0 \quad \Rightarrow 4 g-14 f+c+53=0
$
Also it passes through $(8,1)$
$
\therefore 64+1+16 g+2 f+c=0 \quad \Rightarrow 16 g+2 f+c+65=0
$
Subtracting (3) from (2), we get
$
-2 g+14 f-52=0 \quad \Rightarrow g-7 f+26=0
$
Subtracting (4) from (3), we get
$
-12 g-16 f-12=0 \quad \Rightarrow 3 g+4 f+3=0
$
Solving (5) and (6), we get
Substituting $\mathrm{g}=-5, f=3$ and $c=9$ in (1), we get
$
x^2+y^2-10 x+6 y+9=0
$
Question 6.
Find the equation of a circle which passes through the points $(1,-2)$ and $(4,-3)$ and whose centre lies on the line $3 x+4 y=0$
Solution:
Let the equation of the circle be
$
x^2+y^2+2 g x+2 f y+c=0 \ldots \text { (1) }
$
Since it passes through $(1,-2)$ and $(4,-3)$
$
\therefore 5+2 \mathrm{~g}-4 \mathrm{f}+\mathrm{c}=0 \text {... (2) }
$
and $25+8 g-6 f+c=0$..(3)
Also $(-g,-f)$ centre of circle (1) lies on $3 x+4 y=7$
$
-3 \mathrm{~g}-4 \mathrm{f}=7 \text {...(4) }
$
Subtracting (2) from (3), we get
$
20+6 g-2 f=0 \text {...(5) }
$
Solving (4) and (5), we get

$
g=-\frac{47}{15}, f=\frac{3}{5}\left(=\frac{9}{15}\right)
$
Substituting these values of $g$ and $f$ in (2)
we get
$
c=\frac{11}{3}\left(=\frac{55}{15}\right)
$
$\therefore$ From (1), we get,
$
\begin{aligned}
& x^2+y^2+\frac{94}{15} x+\frac{18}{15} y+\frac{55}{15}=0 \\
& 15 x^2+15 y^2-94 x+18 y+55=0
\end{aligned}
$
Which is the required equation on the circle.

Question 7.
Find the equation (s) of the circle passing through the points (1, 1) and (2,2) and whose radius is 1.
Solution:
Let the equation of the circle be
$
x^2+y^2+2 g x+2 f y+c=0 \ldots \text { (1) }
$
Since it passes through $(1,1)$
$
\begin{aligned}
& \therefore 1+1+2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}=0 \\
& \Rightarrow 2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}+2=0
\end{aligned}
$
Again it passes through $(2,2)$
$
\begin{aligned}
& \therefore 4+4+4 \mathrm{~g}+4 /+\mathrm{c}=0 \\
& \Rightarrow 4 \mathrm{~g}+4 \mathrm{f}+\mathrm{c}+8=0
\end{aligned}
$
Also radius $=\sqrt{g^2+f^2-c}=1$
$\Rightarrow \quad g^2+f^2-c=1$
Subtracting (3) from (2), we get
$
\Rightarrow \quad g+f+3=0 \Rightarrow f=-g-3
$
Adding (2) and (4)
$
\begin{aligned}
& \mathrm{g}^2+2 \mathrm{~g}+\mathrm{f}^2+2 \mathrm{f}+2=1 \\
& \mathrm{~g}^2+2 \mathrm{~g}+(-\mathrm{g}-3)^2+2(-\mathrm{g}-3)+1=0 \\
& \Rightarrow 2 \mathrm{~g}^2+6 \mathrm{~g}+4=0 \Rightarrow \mathrm{g}^2+3 \mathrm{~g}+2=0 \\
& \Rightarrow(\mathrm{g}+1)(\mathrm{g}+2)=0 \Rightarrow \mathrm{g}=-1,-2
\end{aligned}
$
when $g=-1$, from (5), $f=-(-1)-3=1-3=-2$
when $g=-2$, from $(5), \mathrm{f}=-(-2)-3=2-3=-1$
Substituting $g=-1, f=-2$ in (2), we get
$
\begin{aligned}
& 2(-1)+2(-2)+c+2=0 \\
& \Rightarrow-6+c+2=0 \Rightarrow \mathrm{c}=4
\end{aligned}
$
Now putting $g=-1, f=-2$ and $c=4$ in (1), we get
$
x^2+y^2-2 x-4 y+4=0
$
Again putting $g=-2, f=-1$ in (2), we get
$
\begin{aligned}
& 2(-2)+2(-1)+c+2=0 \\
& \Rightarrow-4-2+c+2=0 \Rightarrow c=4
\end{aligned}
$
Putting $g=-2, f=-1$ and $c=4$ (in) (1), we get
$
x^2+y^2-4 x-2 y+4=0
$
Hence the required equation (s) of the circle are
$
x^2+y^2-2 x-4 y+4=0 \text { and } x^2+y^2-4 x-2 y+4=0
$