Circle | Part-3 | S N Dey Examhoop

In the previous article , we discussed Very Short Answer Type Questions. In this article, we will discuss 10 Short Answer type Questions from Chhaya Mathematics , Class 11 (S N De book ).

Short Answer Type Questions of Circle, S N Dey Mathematics, Class 11

1. Find the equation of the circle whose centre is $~(2,-4)~$ and which passes through the centre of the circle $~x^2+y^2-2x+2y-36=0.$

Solution.

The equation of the circle with centre $~(\alpha,\beta)~$ and radius $r$ can be written as :

$~(x-\alpha)^2+(y-\beta)^2=r^2 \rightarrow(1)$

By the given question, $~\alpha=2, ~~\beta=-4.$ Putting these values of $~\alpha,~\beta~$ in $~(1),~$ we get from $~(1),$

$~(x-2)^2+(y+4)^2=r^2 \rightarrow(2)$

Now, comparing the given the circle $~x^2+y^2-2x+2y-36=0,~~$ with the general equation of circle $~x^2+y^2+2gx+2fy+c=0,~$ we get ,

$~2g=-2 \Rightarrow g=-\frac 22=-1,~~2f=2 \Rightarrow f=\frac 22=1.$

Hence, the centre of the given circle is : $(-g,-f)=(1,-1).$

Now, since the circle as represented by $~(2),~$ passes through $~(1,-1),~$ we get,

$~(1-2)^2+(-1+4)^2=r^2 \\ \text{or,}~~ (-1)^2+3^2=r^2 \\ \therefore r^2=1+9=10\rightarrow(3)$

Hence, from $~(2),~(3)~$ we get the required equation of circle which is :

$(x-2)^2+(y+4)^2=10 \\ \text{or,}~~ x^2-2 \cdot x \cdot 2+2^2+y^2+2 \cdot y \cdot 4+4^2=10 \\ \text{or,}~~ x^2+y^2-4x+8y+10=0.$

2. Find the equation of the circle concentric with the circle $~x^2+y^2-4x+6y+4=0~$ and passing through the point $~(2,-2).$

Solution.

Now, comparing the given the circle $~x^2+y^2-4x+6y+4=0~~$ with the general equation of circle $~x^2+y^2+2gx+2fy+c=0,~$ we get ,

$~2g=-4 \Rightarrow g=-\frac 42=-2,~~2f=6 \Rightarrow f=\frac 62=3.$

So, the centre of the circle $~x^2+y^2-4x+6y+4=0~~$ , is : $~(-g,-f)=(2,-3).$

Also, the equation of the circle with the centre $~(2,-3)~$ is given by

$~(x-2)^2+(y+3)^2=r^2 \rightarrow(1)$

Since the circle represented by $~(1),~$ passes through the point $~(2,-2),~$ we get,

$(2-2)^2+(-2+3)^2=r^2 \\ \text{or,}~~ 0+1^2=r^2 \\ \text{or,}~~ r^2=1 \rightarrow(2).$

Hence, from $~(1),~(2)~$ we get the required the equation of the circle which is

$(x-2)^2+(y+3)^2=1 \\ \text{or,}~~ x^2-2 \cdot x \cdot 2+2^2+y^2+2 \cdot y \cdot 3 +3^2-1=0 \\ \therefore x^2+y^2-4x+6y+12=0 .$

Hence, the equation of the circle concentric with the circle $~x^2+y^2-4x+6y+4=0~$ and passing through the point $~(2,-2),$ is $~~x^2+y^2-4x+6y+12=0.$

3. Find the equation of the circle concentric with the circle $~x^2+y^2+4x-6y-13=0~$ and passing through the centre of the circle $~~x^2+y^2-8x-10y-8=0.$

Solution.

Comparing the given the circle $~x^2+y^2+4x-6y-13=0~~$ with the general equation of circle $~x^2+y^2+2gx+2fy+c=0,~$ we get ,

$2g=4 \Rightarrow g=\frac 42=2,~~ 2f=-6 \Rightarrow f=-\frac 62=-3.$

Since we need to find the equation of the circle concentric with the circle $~x^2+y^2+4x-6y-13=0~$ i.e. with the centre $~(-g,-f)=(-2,3)~~$ , the equation of the circle with radius $~r~$ can be written as $~(x+2)^2+(y-3)^2=r^2 \rightarrow(1)$

Now, we have to find the centre of the circle $~~x^2+y^2-8x-10y-8=0.$

Comparing the given the circle $~x^2+y^2-8x-10y-8=0~~$ with the general equation of circle $~x^2+y^2+2gx+2fy+c=0,~$ we get ,

$2g=-8 \Rightarrow g=-\frac 82=-4,~~ 2f=-10 \Rightarrow f= -\frac{10}{2}=-5.$

So, the centre of the circle : $~(-g,-f)=(4,5).$

Now, the circle represented by $~(1),~$ passes through the point $~(4,5),$ so

$(4+2)^2+(5-3)^2=r^2 \\ \Rightarrow r^2=6^2+2^2=40\rightarrow(2) .$

Finally, from $~(1),~(2)~$ we get the equation of the circle which is

$(x+2)^2+(y-3)^2=40 \\ \text{or,}~~ x^2+2 \cdot x \cdot 2+2^2+y^2-2 \cdot y \cdot 3 +3^2-40=0 \\ \therefore x^2+y^2+4x-6y-27=0.$

Hence, the required equation of the circle is $~~x^2+y^2+4x-6y-27=0.$

4. Find the equation of the straight line which passes through the centre of the circle $~~x^2+y^2+2x+2y-23=0~$ and is perpendicular to the straight line $~x-y+8=0.$

Solution.

The equation of the straight line which is perpendicular to the straight line $~x-y+8=0,~$ can be written as : $~x+y+k=0,~~k~$ being an arbitrary constant.

Now, we need to find the centre of the circle $~~x^2+y^2+2x+2y-23=0.~$

Comparing the given the circle $~x^2+y^2-8x-10y-8=0~~$ with the general equation of circle $~x^2+y^2+2gx+2fy+c=0,~$ we get ,

$~2g=2 \Rightarrow g=\frac 22=1,\~~ 2f=2 \Rightarrow f=\frac 22=1.$

Hence, the centre of the circle $~x^2+y^2+2x+2y-23=0,~$ is given by $~~(-g,-f)=(-1,-1).$

Now, since the straight line $~x+y+k=0,~$ passes through the point $~(-1,-1),~$ we have

$-1-1+k=0 \Rightarrow k=2$

Finally putting the value of $~k=2~$ in $~x+y+k=0~$ , we get the required equation of the straight line which is $~~x+y+2=0.$

5. Find the equation of each of the circles passing through the points

$(i)~(0,0),(1,2),(2,0) ~~(ii)~~ (2,-1),~(2,3),~(4,-1)$

Solution.

We know that the general form of the equation of a circle is

$~x^2+y^2+2gx+2fy+c=0 \rightarrow(1)$

Since the circle in $~(1),~$ passes through the points $~(0,0),(1,2),(2,0),~$ we get

$0+0+0+0+c=0 \Rightarrow c=0 , \\ ~ 1^2+2^2+2g \times 1+2f \times 2+0=0 \\ \text{or,}~~ 5+2g+4f=0 \rightarrow(2) \\ ~ 2^2+0+2g \times 2+2f \times 0+0=0 \\ \text{or,}~~ 4+4g=0 \\ \text{or,}~~ g=-\frac 44=-1.$

Now, putting the value of $~g~$ in $~(2),~$ we get,

$~5+2 \times (-1)+4f=0 \\ \text{or,}~~3+4f=0 \\ \text{or,}~~ f=-\frac 34.$

Hence, by putting the values of $~c,g,f~$ in $(1),$ we get the required equation of circle which is

$x^2+y^2+2(-1) \cdot x+2(-3/4) \cdot y+0=0 \\ \text{or,}~~ x^2+y^2-2x-\frac 32y=0 \\ \text{or,}~~ 2x^2+2y^2-4x-3y=0.$

6.Show that the points $~(2,0),~(5,-3),~(2,-6)~$ and $~(-1,-3)~$ are concylic ; find the equation of the circle on which the points lie and the co-ordinates of the centre of the circle.

Solution.

The general equation of circle $~x^2+y^2+2gx+2fy+c=0 \rightarrow(1).$

Since the points $~(2,0),~(5,-3),~(2,-6)~$ and $~(-1,-3)~$ are concylic, we get from $~(1),$

$2^2+0+2g \times 2+0+c=0\\ \Rightarrow 4+4g+c=0\rightarrow(2)$

$5^2+(-3)^2+2g \times 5+2f \times (-3)+c=0 \\ \Rightarrow 25+9+10f-6f+c=0 \\ \Rightarrow 34+10g-6f+c=0 \rightarrow(3)$

$2^2+(-6)^2+2g \times 2+2f \times (-6)+c=0 \\ \Rightarrow 4+36+4g-12f+c=0 \\ \Rightarrow 40+4g-12f+c=0 \rightarrow(4)$

$(-1)^2+(-3)^2+2g \times (-1)+2f \times (-3)+c=0 \\ \Rightarrow 1+9-2g-6f+c=0 \\ \Rightarrow 10-2g-6f+c=0 \rightarrow(5)$

Subtracting $~(2)~$ from $~(3),~$ we get

$30+6g-6f=0 \\ \text{or,}~~6(5+g-f)=0 \\ \text{or,}~~5+g-f=0\rightarrow(6)$

Subtracting $~(3)~$ from $~(4),~$ we get

$6-6g-6f=0 \\ \text{or,}~~6(1-g-f)=0 \\ \text{or,}~~ 1-g-f=0\rightarrow(7)$

From $~(6),~(7)~$ , we get

$(5+g-f)+(1-g-f)=0 \\ \text{or,}~~ 6-2f=0 \\ \text{or,}~~ f=\frac 62=3 \rightarrow(8)$

From $~(7),~(8)~$ we get,

$~1-g-3=0 \Rightarrow g=-2 \rightarrow(9)$

Putting the value of $~f,g~$ in $~(5),~$ we get,

$~10 -2\times (-2)-6\times 3+c=0 \Rightarrow c=4.$

Hence, the equation of the required circle is

$x^2+y^2+2 \times (-2) \times x+2 \times 3 \times y+4=0 \\ \text{or,}~~ x^2+y^2-4x+6y+4=0 \rightarrow(10)$

Comparing $~x^2+y^2-4x+6y+4=0~~$ with the general equation of circle $~x^2+y^2+2gx+2fy+c=0,~$ we get ,

$2g=-4 \Rightarrow g=-\frac 42=-2 , ~~ 2f=6 \Rightarrow f=\frac 62=3.$

Hence, the co-ordinates of the centre of the circle is $~(-g,-f)=(2,-3).$

7. Prove that the centres of the circles $~x^2+y^2-10x+9=0, ~ x^2+y^2-6x+2y+1=0$ and $~~x^2+y^2-18x-4y+21=0~$ lie on a line; find the equation of the line on which they lie.

Solution.

We know that the general form of the equation of a circle is $~x^2+y^2+2gx+2fy+c=0 \rightarrow(1)$

The centre of the circle represented by $~(1),~$ is $~(-g,-f).$

Now, comparing $~x^2+y^2-10x+9=0 \rightarrow(2)~$ with $~(1),~$ we get,

$2g=-10 \Rightarrow g=-\frac{10}{2}=-5,~2f=0 \Rightarrow f=0.$

So, the centre of the circle represented by $~(2),~$ is $~(-g,-f)=(5,0).$

Similarly, the centre of the circle $~x^2+y^2-6x+2y+1=0~$ is $~(3,-1)~$ and the centre of the circle $~x^2+y^2-18x-4y+21=0~$ is $~(9,2).$

Now, the equation of the line passing through $~(5,0),~(3,-1)~~$ is :

$\frac{y-0}{x-5}=\frac{-1-0}{3-5} \\ \text{or,}~~ \frac{y}{x-5}=\frac 12 \\ \text{or,}~~ x-2y=5 \rightarrow(3).$

Now, the putting $~(9,2)~$ in $~(3)~$ we get, $~ 9- 2 \times 2=5.$

Hence, the point $~(9,2)~$ satisfies the equation $~(3).$

So, the centres of three circles lie on the line $~x-2y=5.$

8. Show that the centres of the following circles lie on a line and their radii are in A.P. : $~ x^2+y^2=1,~ x^2+y^2+6x-2y-6=0,~x^2+y^2-12x+4y-9=0.$

Solution.

The equation of the first circle can be written as $~(x-0)^2+(y-0)^2=1^2~$ so that we can say the centre of the first circle is $~(0,0)~$ and radius is $~1~$ unit.

The equation of the second circle is $~x^2+y^2+6x-2y-6=0~$ so that the the centre of the second circle is :

$(-g,-f)=\left(-\frac 62,\frac 22\right)=(-3,1).$

Also, the radius is :

$\sqrt{g^2+f^2-c}=\sqrt{3^2+(-1)^2+6}=\sqrt{16}=4 ~~\text{unit}$

The equation of the third circle is $~x^2+y^2-12x+4y-9=0~$ so that the the centre of the third circle is :

$(-g,-f)=\left(\frac{12}{2},-\frac 42\right)=(6,-2)$

Also, the radius is :

$\sqrt{g^2+f^2-c}=\sqrt{(-6)^2+2^2+9}=\sqrt{49}=7~~\text{unit.}$

Clearly, the radii of three circles are $~1~\text{unit},4~\text{unit},7~\text{unit}~$ respectively and so $~4-1=3=7-4~$ and so the radii are in A.P.

Again, the slope of the straight line joining the points $~(0,0)~$ and $~(-3,1)~$ is : $=\frac{1-0}{-3-0}=-\frac 13 \rightarrow(1).$

Also, the slope of the straight line joining the points $~(-3,1)~$ and $~(6,-2)~$ is : $=\frac{-2-1}{6+3}=-\frac 13\rightarrow(2).$

Hence, from $~(1)~$ and $~(2),~$ we can conclude that the centres of the following circles lie on a line.

9. Find the equation of the circle which passes through the origin and cuts off intercepts $~3~$ unit and $~4~$ unit from $~x~$ and $~y$ -axes respectively. Find the equation of that diameter of the circle which passes through the origin.

Solution.

Suppose that the circle which passes through the origin , cuts off intercepts $~3~$ unit and $~4~$ unit from $~x~$ and $~y$ -axes at the point $~A~$ and $~B~$ respectively. Then, $~A \equiv (3,0),~B \equiv (0,4).$

The circle passes through the origin and $~\angle{AOB}=90^{\circ}.$

$\because~~AB~$ is the diameter of the circle, the equation of the circle having diameter $~AB~$ with $~A \equiv (3,0),~B \equiv (0,4),~$ can be written as

$(x-3)(x-0)+(y-0)(y-4)=0 \\ \text{or,}~~ x^2+y^2-3x-4y=0.$

Now , we need to find the equation of the diameter of this circle which passes through the $~(0,0)$ .

The centre of the circle is $~~\left(\frac{3+0}{2},\frac{0+4}{2}\right)=(3/2,2).$

Now, the diameter of this circle passes through $~(0,0),~(3/2,2).$

So, the equation of the straight line which passes through $~(0,0),~(3/2,2)~$ is given by :

$\frac{y-0}{x-0}=\frac{2-0}{\frac 32-0} \\ \text{or,}~~ 4x-3y=0.$

10. Find the equation of the circle circumscribing the triangle formed by the straight line $~2x+3y=6~$ with the axes of co-ordinates. What is the diameter of the circle ?

Solution.

Let the given straight line intercepts the axes of co-ordinates at the points $~A~$ and $~B~$ respectively.

Now, $~2x+3y=6 \Rightarrow \frac x3+\frac y2=1 \rightarrow(1).$

From $~(1),~$ we notice that the straight line cuts off $~3~$ units from the $~x-$ axis and $~2-$ units from $~y-$ axis.

$\therefore~ A \equiv (3,0),~~ B \equiv (0,2).$

Now, $~\angle{AOB}=90^{\circ}.~~$ Clearly, $\Delta AOB~$ is a right angled triangle and $~C~$ is the mid point of $~AB~$ which is a diameter of circumscribed circle with extremities $~(3,0),~(0,2).$