Circle | Part-4 | S N Dey Examhoop

In the previous article , we discussed Very Short Answer Type Questions. In this article, we will discuss 10 more 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

11. The extremities of a diameter of a circle are the points $~(4,-2)~$ and $~(-1,3)~$ ; Find the equation of the circle. Also find the equation of that diameter of this circle which passes through the origin.

Solution.

The equation of any circle having the extremities $~(4,-2)~$ and $~(-1,3)~$ of a diameter is given by

$(x-4)(x+1)+(y+2)(y-3)=0 \\ \text{or,}~~ x^2+y^2-3x-y-10=0.$

Now, we need to find the equation of that diameter of this circle which passes through the origin.

The centre of the circle is $~~\left(\frac{4-1}{2},\frac{-2+3}{2}\right)=(3/2,1/2).$

Hence equation of the diameter that passes through $~(0,0)~$ and $~(3/2,1/2)~$ is given by :

$\frac{y-0}{x-0}=\frac{\frac 12-0}{\frac 32-0} \Rightarrow x-3y=0.$

12. Find the equation of the circle through the points $~(4,3)~$ and $~(-2,5)~$ and having its centre on the line $~~2x-3y=4.$

Solution.

Let $~(\alpha,\beta)~$ be the centre of the circle.

Since $~(\alpha,\beta)~$ lies on the line $~~2x-3y=4,~$ $~ 2\alpha-3\beta=4 \rightarrow(1).$

Again, since the circle passes through the points $~(4,3)~$ and $~(-2,5)~$ , the distance of $~(\alpha,\beta)~$ from the points $~(4,3)~$ and $~(-2,5)~$ , will be same.

$\therefore ~\sqrt{(\alpha+2)^2+(\beta-5)^2}=\sqrt{(\alpha-4)^2+(\beta-3)^2} \\ \text{or,}~~ (\alpha+2)^2+(\beta-5)^2=(\alpha-4)^2+(\beta-3)^2 \\ \text{or,}~~\alpha^2+4\alpha+4+\beta^2-10\beta+25=\alpha^2-8\alpha+16+\beta^2-6\beta+9 \\ \text{or,}~~ 12 \alpha-4\beta=-4 \\ \text{or,}~~ 3\alpha-\beta=-1 \rightarrow(2)$

Now, solving $~(1)~$ and $~(2),~$ we get, $~ \alpha=-1,~\beta=-2.$

So, the centre of the circle is $~(-1,-2)~$ and the radius of the circle is

$=\sqrt{(-2+1)^2+(5+2)^2}=\sqrt{50}.$

Hence, the equation of the circle is

$(x+1)^2+(y+2)^2=(\sqrt{50})^2 \\ \text{or,}~~ x^2+y^2+2x+4y-45=0.$

13. A circle has its centre on the straight line $~5x-2y+1=0~$ and cuts off the $~x-$ axis at the two points whose abscissae $~(-5)~$ and $~3;~$ find the equation of the circle and its radius.

Solution.

The general form of the equation of the circle can be written as

$x^2+y^2+2gx+2fy+c=0 \rightarrow (1)~$ where $~(-g,-f)~$ is the centre of the circle and the radius $~~(r)=\sqrt{g^2+f^2-c} \rightarrow(2)$

According to the problem, the circle passes through the points $~(-5,0)~$ and $~(3,0).$

So, by $~(1),~$ we get

$~(-5)^2+0^2+2g \times (-5)+2f \times 0+c=0 \\ \text{or,}~~ 25-10g+c=0 \rightarrow(3)$

Similarly, by $~(1),~$ we get

$~3^2+0^2+2g \times 3+2f \times 0+c=0 \\ \text{or,}~~ 9+6g+c=0 \rightarrow(4)$

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

$~16-16g=0 \Rightarrow g=1$

Since the circle has its centre on the straight line $~5x-2y+1=0~$ , so
$~-5g+2f+1=0 \rightarrow(5)$

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

$~-5 \times 1+2f+1=0 \Rightarrow f=\frac 42=2.$

Now, putting the value of $~g,f~$ in $~(4),~$ we get

$~9+6 \times 1+c=0 \Rightarrow c=-15.$

By $~(2),~$ we get the radius $~(r)=\sqrt{1^2+2^2+15}=\sqrt{20}=2\sqrt{5}~\text{unit}$

By $~(1),~$ we get the equation of the circle as

$~x^2+y^2+2 \times 1 \times x+2 \times 2 \times y-15=0 \\ \text{or,}~~ x^2+y^2+2x+4y-15=0.$

14. The equation of a diameter of a circle is $~2x-y+4=0~$ and it passes through the points $~(4,6)~$ and $~(1,9).$ Find the equation of the circle , the co-ordinates of its centre and length of its radius.

Solution.

Let $~(\alpha,\beta)~$ be the centre of the circle so that $~2\alpha-\beta+4=0 \rightarrow(1)~$ as the diameter of the circle passes through the centre of the circle.

Again, the circle passes through the points $~(4,6)~$ and $~(1,9),~$ so the centre $~(\alpha,\beta)~$ is equidistant from the points $~(4,6)~$ and $~(1,9).$

$\therefore~\sqrt{(\alpha-4)^2+(\beta-6)^2}=\sqrt{(\alpha-1)^2+(\beta-9)^2} \\ \text{or,}~~\alpha^2-8\alpha+16+\beta^2-12\beta+36\\~~=\alpha^2-2\alpha+1+\beta^2-18\beta++81. \\ \text{or,}~~ 6\alpha-6\beta+30=0 \\ \therefore \alpha-\beta+5=0 \rightarrow(2)$

Solving $~(1),~(2)~$ we get, $~\alpha=1,~~ \beta=6.$

So, the centre of the circle is $~(1,6)~$ and radius of the circle is $~\sqrt{(4-1)^2+(6-6)^2}=3~\text{unit.}$

$\therefore~$ the equation of the circle

$~(x-1)^2+(y-6)^2=3^2 \\ \text{or,}~~x^2+y^2-2x-12y+28=0.$

15. A circle passes through through the points $~(-6,5),~(-3,-4)~$ and its radius is $~5~$ unit ; find the equation of the circle.

Solution.

Let the centre of the circle be $~(\alpha,\beta).$

$\therefore~$ the equation of the circle can be written as

$~(x-\alpha)^2+(y-\beta)^2=5^2 \rightarrow(1)~~$ which passes through the points $~(-6,5)~$ and $~(-3,-4).$

$~\therefore~(-6-\alpha)^2+(5-\beta)^2=5^2 \\ \text{or,}~~ \alpha^2+\beta^2+12\alpha-10\beta+36=0\rightarrow(2)$

$\text{Also,}~(-3-\alpha)^2+(-4-\beta)^2=5^2 \\ \text{or,}~~ \alpha^2+\beta^2+6\alpha+8\beta=0 \rightarrow(3)$

Now, subtracting $~(3)~$ from $~(2)~$ we get,

$6\alpha-18\beta+36=0 \\ \text{or,}~~ 6(\alpha-3\beta+6)=0 \\ \text{or,}~~\alpha=3 \beta-6 \rightarrow(4)$

By using $~(2),~(4)~$ we get,

$(3\beta-6)^2+\beta^2+12(3\beta-6)-10\beta+36=0 \\ \text{or,}~~ 10\beta^2-10\beta=0 \\ \text{or,}~~ 10\beta(\beta-1)=0 \\ \therefore~\beta=0,1.$

$~\therefore~\alpha=(3 \times 0-6), (3 \times 1-6)~=-6,-3.$

So, the required equation of the circle is

$(x+6)^2+(y-0)^2=5^2~$ and $~(x+3)^2+(y-1)^2=5^2 ~$

or, $~x^2+y^2+12x+11=0,~x^2+y^2+6x-2y-15=0.$

$16.~~ 3x+y=5~$ and $~x+y+1=0~$ are two diameters to the circle which passes through the point $~(-2,2)~$ . Find its equation. Also find the radius of the circle.

Solution.

Clearly, the centre of the circle will be the intersection of two given diameters.

$\text{Now,}~3x+y=5 \rightarrow(1)~$ and $~x+y+1=0 \rightarrow(2)~$ are two given diameters.

Solving $~(1),~(2)~$ we get, $~x=3,y=-4$

So, the centre of the circle is : $~(3,-4).$ Also, it is given that the circle passes through the point $~(-2,2)~$ .

Hence, the radius $(r)=\sqrt{(3+2)^2+(-4-2)^2}~\text{unit}=\sqrt{61}~\text{unit}.$

Finally, the equation of the circle is

$(x-3)^2+(y+4)^2=(\sqrt{61})^2 \\ \text{or,}~~ x^2-6x+9+y^2+8y+16=61 \\ \text{or,}~~ x^2+y^2-6x+8y-36=0.$

17. A circle passes through the points $~(-3,4),~(1,0)~$ and its centre lies on the $~x-$ axis. Find the equation of the circle.

Solution.

Let the centre of the circle $~(\alpha,0),~$ because the centre lies on the $~x-$ axis. Since the circle passes through the points $~(-3,4),~(1,0)~$ , the points are equidistant from the centre of the circle.

$\therefore~ \sqrt{(\alpha+3)^2+(0-4)^2}=\sqrt{(\alpha-1)^2+(0-0)^2} \\ \text{or,}~~ \alpha^2+6\alpha+9+16=\alpha^2-2\alpha+1 \\ \text{or,}~~ 8\alpha=-24 \\ \therefore \alpha=-3.$

$~\therefore~$ the centre of the circle $~(-3,0).$

$~\therefore~$ the radius of the circle $~\sqrt{(-3-1)^2+(0-0)^2}=4~\text{unit}$

$\therefore~$ the equation of the circle is

$(x+3)^2+(y-0)^2=4^2 \ \text{or,}~~ x^2+6x+9+y^2-16=0 \ \therefore~ x^2+y^2+6x-7=0.$

18. Find the equation of the circle which has its centre on the line $~y=2~$ and which passes through the points $~(2,0)~$ and $~(4,0).$

Solution.

Let the centre of the circle be $~(\alpha,2)~$ as its centre lie on the line $~y=2.$

Since the circle passes through the points $~(2,0),~(4,0)~$ , the points are equidistant from the centre of the circle.

$\therefore~\sqrt{(\alpha-2)^2+(2-0)^2}=\sqrt{(\alpha-4)^2+(2-0)^2} \\ \text{or,}~~\alpha^2-4\alpha+4+4=\alpha^2-8\alpha+16+4 \\ \text{or,}~~8\alpha-4\alpha=16-4 \\ \text{or,}~~\alpha=\frac{12}{4}=3.$

$\therefore~$ the centre of the circle $~(3,2)~$ and radius $=\sqrt{(3-2)^2+(2-0)^2}=\sqrt{5}~~\text{unit}.$

$\therefore~$ the equation of the circle is

$(x-3)^2+(y-2)^2=(\sqrt{5})^2 \\ \text{or,}~~ x^2-6x+9+y^2-4y+4=5 \\ \text{or,}~~ x^2+y^2-6x-4y+8=0.$

19. Find the equation to the circle which touches the $~y-$ axis at the origin and passes through the point $~(\alpha,\beta).$

Solution.

Since the circle which touches the $~y-$ axis at the origin, from the following figure we notice that the centre $~(C)~$ of the circle lies on the $~x-$ axis.

Let the radius of the circle $~(OC)=r~$ and the co-ordinate of the centre of the circle $~(C) \equiv (r,0).$ Now, since the circle passes through the point $~(\alpha,\beta),~$ the radius of the circle is the distance between $~C(r,0)~$ and $~(\alpha,\beta).$

$\therefore~ r=\sqrt{(r-\alpha)^2+(0-\beta)^2} \\ \text{or,}~~ r^2=r^2-2r\alpha+\alpha^2+\beta^2 \\ \text{or,}~~r=\frac{\alpha^2+\beta^2}{2\alpha}\rightarrow(1).$

Now, the equation of the circle with centre $~(r,0)~$ and radius $~r~$ is

$(x-r)^2+y^2=r^2 \\ \text{or,}~~ x^2-2rx+r^2+y^2=r^2 \\ \text{or,}~~x^2+y^2-2rx=0 \\ \text{or,}~~ x^2+y^2-\frac{2x(\alpha^2+\beta^2)}{2\alpha}=0 ~~[\text{By (1)}] \\ \text{or,}~~\alpha(x^2+y^2)=x(\alpha^2+\beta^2).$

20. A circle touches the $~x-$ axis at $~(3,0)~$ and its radius is twice the radius of the circle $~x^2+y^2-2x-2y-2=0;~$ find the equation of the circle and the length of its chord intercepted on the $~y-$ axis.

Solution.

We can rearrange the given circle as

$x^2+y^2-2x-2y-2=0 \\ \text{or,}~~ (x-1)^2+(y-1)^2=4=2^2\rightarrow(1)$

So, the radius of the circle as represented by $~(1),~$ is $~2~$ unit. Our circle of concern will have a radius of $~2 \times 2=4~$ units. Now, suppose that the centre of circle of concern is $~(\alpha,4)~$ so that the equation of circle can be written as