Circle | Part-7 | S N Dey Examhoop

In the previous article , we have completed the discussions about the Short Answer Type Questions. In this article, we will discuss 10 Long Answer type Questions from Chhaya Mathematics , Class 11 (S N De book ).

Circle(Long Answer Type Questions), S N Dey Solution Class 11

Long Answer Type Questions (1-10) | Circle | S N Dey Solution

1.Find the equation to the circles which touch the $~y-$ axis and pass through $~(-2,1)~$ and $~(-4,3).$

Solution.

The equation of the circle touching the $~y-$ axis can be written as

$~(x-a)^2+(y-\beta)^2=a^2~$ where $~(a,\beta)~$ is the centre of the circle with radius $~a~$ unit.

Since the circle passes through $~(-2,1)~$ ,

$(-2-a)^2+(1-\beta)^2=a^2 \\ \text{or,}~~(2+a)^2+(1-\beta)^2=a^2 \\ \text{or,}~~4+4a+a^2+1-2\beta+\beta^2=a^2 \\ \text{or,}~~ \beta^2+4a-2\beta+5=0 \rightarrow(1)$

Again, since the circle passes through $~(-4,3)~$ ,

$(-4-a)^2+(3-\beta)^2=a^2 \\ \text{or,}~~ (4+a)^2+(3-\beta)^2=a^2 \\ \text{or,}~~16+8a+a^2+9-6\beta+\beta^2=a^2 \\ \text{or,}~~ \beta^2+8a-6\beta+25=0\rightarrow(2)$

Subtracting $~(1)~$ from $~(2),~$ we get

$4a-4\beta+20=0 \\ \text{or,}~~ 4(a-\beta+5)=0 \\ \therefore a=\beta-5 \rightarrow(3)$

Now, from $~(1)~$ and $~(2)~$ we get,

$\beta^2+4(\beta-5)-2\beta+5=0 \\ \text{or,} ~~ \beta^2+4\beta-20-2\beta+5=0 \\ \text{or,} ~~ \beta^2+2\beta-15=0 \\ \text{or,} ~~ \beta^2+5\beta-3\beta-15=0 \\ \text{or,} ~~\beta(\beta+5)-3(\beta+5)=0 \\ \text{or,} ~~ (\beta+5)(\beta-3)=0 \\ \text{or,} ~~\beta=-5,3 \rightarrow(4)$

$\text{For}~\beta=-5,3 ~$ , corresponding values of $~a=-10,-2~~[\text{By (3)}].$

So, the required equation of circle (for $~a=-10,~\beta=-5$ ) is

$(x+10)^2+(y+5)^2=(-10)^2 \\ \text{or,}~~ x^2+20x+100+y^2+10y+25=100 \\ \text{or,}~~ x^2+y^2+20x+10y+25=0.$

and also the required equation of circle (for $~a=-2,~\beta=3$ ) is

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

2. If two straight lines $~3x-2y=8~$ and $~2x-y=5~$ lie along two diameters of a circle , which touches the $~x-$ axis, find the equation of the circle.

Solution.

According to the problem, center of the circle will be the point where both lines cross each other. Now, solving two straight lines $~3x-2y=8~$ and $~2x-y=5~$ we get, $~x=2,~y=-1.$

So, the centre of the circle $~(2,-1).$

Since, the circle touches $~x-$ axis, so the equation of the circle is

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

3. A circle touches the lines $~x=0,~ y=0~$ and $~x+y=1.~$ If the centre of the circle lies in the first quadrant , show that there are two such circles and find their equations . Specify which of these is inscribed within the triangle formed by the given lines.

Solution.

Since the circle touches both $~x-$ axis and $~y-$ axis , the centre of the circle is $~(a, a).$ So, the equation of the circle is $~(x-a)^2+(y-a)^2=a^2 \rightarrow(1).$

Since $~x+y=1~$ is a tangent to the circle $~(1),~$ so the distance of $~x+y=1~$ from the centre $~(a,a)~$ of the circle = the radius of the circle.

$\text{So,}~ \frac{|2a-1|}{\sqrt{1^2+1^2}}=a \\ \text{or,} ~~\frac 12(2a-1)^2=a^2 \\ \text{or,} ~~ 4a^2-4a+1=2a^2 \\ \text{or,} ~~ 2a^2-4a+1=0 \\ \text{or,} ~~ a=\frac{4 \pm \sqrt{(-4)^2-4 \times 2 \times 1}}{2 \times 2}=\frac 14(4 \pm \sqrt{8}) \\ \text{or,} ~~ a=\frac 12(2 \pm \sqrt{2})$

So, the equation of the circle $~ (x-a)^2+(y-a)^2=a^2~$ where $~a=\frac 12(2 \pm \sqrt{2}).$

Now, since the points $~\left(\frac12(2-\sqrt{2}),\frac12(2-\sqrt{2} )\right)~$ and $~(0,0)~$ lie on the same side of the straight line $~x+y=1~$ , the circle with radius $~\frac 12(2-\sqrt{2})~$ unit is inscribed within the triangle formed by the given lines.

4. Find the equations to the circles which which touch the axis of $~y-$ at a distance $~+4~$ from the origin and intercept a length $~6~$ unit on the axis of $~x.$

Solution.

Let the equation of the circle be $~(x-a)^2+(y-4)^2=a^2 \rightarrow(1).$

$\text{By (1), we get,}$

$x^2-2ax+a^2+y^2-8y+16=a^2 \\ \text{or,}~~ x^2+y^2-2ax-8y+16=0\rightarrow(2).$

Now, the the length of the intercept on $~x-$ axis

$2\sqrt{a^2-16}=6 \\ \text{or,}~~ a^2-16=(6/2)^2=9 \\ \text{or,}~~ a^2=9+16 \\ \text{or,}~~ a=\pm \sqrt{25}=\pm 5 \rightarrow(3).$

By $~(3),$ we get two circles by two corresponding values of $~a.$

$\therefore~$ Using $~(2),~(3)~$ two circles are given by

$x^2+y^2-2x (\pm 5)-8y+16=0 \\ \text{or,} ~~ x^2+y^2 \pm 10x-8y+16=0.$

5. A circle passes through the point $~(-2,1)~$ and touches the straight line $~3x-2y=6~$ at the point $~(4,3)~$ . Find its equation.

Solution.

6. Find the equation of the circle which touches the $~x-$ axis at a distance $~+5~$ unit from the origin and cuts off an intercept of length $~24~$ unit from the $~y-$ axis.

Solution.

Let the equation of the circle be $~(x-5)^2+(y-a)^2=a^2\rightarrow(1)$

The circle $~(1)~$ can be written as

$x^2-10x+5^2+y^2-2ay+a^2=a^2 \\ \text{or,}~~ x^2+y^2-10x-2ay+25=0\rightarrow(2)$

The length of the intercept on $~y-$ axis

$2\sqrt{a^2-25}=24 \\ \text{or,}~~ a^2-25=(24/2)^2 \\ \text{or,}~~a^2=144+25 \\ \text{or,}~~ a=\pm\sqrt{169}=\pm 13 \rightarrow(3).$

So, by $~(2)~$ and $~(3)~$ we get the equation of the circle as follows :

$~x^2+y^2-10x \pm26y+25=0.$

7. Show that the circles $~x^2+y^2-4x+6y+8=0~$ and $~x^2+y^2-10x-6y+14=0~$ touch each other externally ; find the co-ordinates of their point of contact.

Solution.

Two given circles are

$~x^2+y^2-4x+6y+8=0 \rightarrow(1),\\ x^2+y^2-10x-6y+14=0\rightarrow(2).$

Comparing $~(1)~$ with $~x^2+y^2+2gx+2fy+c=0~$ we get,

$2g=-4 \Rightarrow -g=4/2=2,\\~~2f=6 \Rightarrow -f=-(6/2)=-3,\\~~c=8\\~\sqrt{g^2+f^2-c}=\sqrt{4+9-8}=\sqrt{5}.$

So, for the circle $~(1),~$ centre $~C_1 \equiv (-g,-f)=(2,-3)~$ and radius $(r_1)=\sqrt{g^2+f^2-c}=\sqrt{5}.$

Comparing $~(2)~$ with $~x^2+y^2+2gx+2fy+c=0~$ we get,

$2g=-10 \Rightarrow -g=10/2=5,\\~~2f=-6 \Rightarrow -f=6/2=3,\\~~c=14\\~\sqrt{g^2+f^2-c}=\sqrt{25+9-14}=\sqrt{20}=2\sqrt{5}.$

So, for the circle $~(2),~$ centre $~C_1 \equiv (-g,-f)=(5,3)~$ and radius $(r_1)=\sqrt{g^2+f^2-c}=2\sqrt{5}.$

Distance ( $~C_1C_2~$ ) between the centres of two circles

$=\sqrt{(5-2)^2+(3+3)^2}\\~~~=\sqrt{9+36}\\~~~=\sqrt{45}\\~~~=\sqrt{5 \times 9}\\~~~=3\sqrt{5}\\~~~=\sqrt{5}+2\sqrt{5}\\~~~=r_1+r_2$

So, two circles touch each other externally.

Again, $~r_1 :r_2=1 :2~~$ so that the co-ordinates of their point of contact is

$\left(\frac{1\times 5+ 2\times 2}{1+2},\frac{1 \times 3+2 \times (-3)}{1+2}\right)=(3,-1).$

8. Prove that the circles $~x^2+y^2+4x-10y-20=0~$ and $~x^2+y^2-4x-4y+4=0~$ touch each other internally. Find the equation of their common tangent.

Solution.

Two given circles are

$x^2+y^2+4x-10y-20=0\rightarrow(1),\\~~~ x^2+y^2-4x-4y+4=0\rightarrow(2).$

Comparing $~(1)~$ with $~x^2+y^2+2gx+2fy+c=0~$ we get,

$2g=4 \Rightarrow -g=-4/2=-2,\\~~~2f=-10 \Rightarrow -f=10/2=5,\\~~~c=-20\\~~~\sqrt{g^2+f^2-c}=\sqrt{4+25+20}=\sqrt{49}=7.$

So, for the circle $~(1),~$ centre $~C_1 \equiv (-g,-f)=(-2,5)~$ and radius $(r_1)=\sqrt{g^2+f^2-c}=7.$

In a similar way, from $~(2)~$ we get, the centre $~C_2 \equiv (-g,-f)=(2,2)~$ and radius $(r_2)=\sqrt{g^2+f^2-c}=\sqrt{4+4-4}=2.$

Distance ( $~C_1C_2~$ ) between the centres of two circles

$\sqrt{(2+2)^2+(2-5)^2}=5=7-2=r_1-r_2\rightarrow(3).$

Hence, by $~(3),~$ we can conclude that two circles touch each other internally.

Now, subtracting $~(2)~$ from $~(1),~$ we get

$8x-6y-24=0 \Rightarrow 4x-3y=12 \rightarrow(4)$

So, the equation in $~(4)~$ , represents the equation of their common tangent.

9. If the circles $~x^2+y^2+2ax+c^2=0~$ and $~x^2+y^2+2by+c^2=0~$ touch each other , prove that , $~\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}.$

Solution.

Two given circles are

$x^2+y^2+2ax+c^2=0 \rightarrow(1),~~ x^2+y^2+2by+c^2=0\rightarrow(2).$

The centre of the circle $~(1)~$ is $~A(-a,0)~$ and the radius $~(r_1)=\sqrt{a^2-c^2}~\text{unit}.$

The centre of the circle $~(2)~$ is $~B(0,-b)~$ and the radius $~(r_2)=\sqrt{b^2-c^2}~\text{unit}.$

$AB=r_1 \pm r_2 \\ \text{or,}~~ \sqrt{a^2+b^2}=\sqrt{a^2-c^2} \pm \sqrt{b^2-c^2} \\ \text{or,}~~ a^2+b^2 =a^2-c^2+b^2-c^2 \pm 2\sqrt{(a^2-c^2)(b^2-c^2)} \\ \text{or,}~~ 2c^2=\pm2\sqrt{a^2b^2-c^2(a^2+b^2)+c^4} \\ \text{or,}~~c^4=a^2b^2-c^2(a^2+b^2)+c^4 \\ \text{or,}~~ c^2(a^2+b^2)=a^2b^2 \\ \text{or,}~~ \frac{a^2+b^2}{a^2b^2}=\frac{1}{c^2} \\ \text{or,}~~\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}~~\text{(proved)}$

10. Prove that the circles $~x^2+y^2-2x-4y-12=0~$ and $~3x^2+3y^2-2x+4y-140=0~$ touch each other. Find the co-ordinates of the point of contact.

Solution.

Two given equation of circles are $~x^2+y^2-2x-4y-12=0\rightarrow(1)~$ and $~3x^2+3y^2-2x+4y-140=0\rightarrow(2).$

Clearly, centre of circle $~(1)~$ is $~A(1,2)~$

and radius $~(r_1)$ is

$=\sqrt{g^2+f^2-c}\\~~=\sqrt{(-1)^2+(-2)^2-(-12)}\\~~=\sqrt{1+4+12}\\~~=\sqrt{17}~\text{unit.}$

Again, the circle $~(2)~$ can be rewritten as $~x^2+y^2-(2/3)x+(4/3)y-\frac{140}{3}=0~\rightarrow(3)$ ans so centre of circle $~(3)~$ is $~\left(\frac 13,-\frac 23\right).$

Also, the radius ( $r_2$ ) of circle of $~(3)~$ is

$=\sqrt{g^2+f^2-c}\=\sqrt{(-1/3)^2+(2/3)^2-(-140/3)}\\~~=\sqrt{\frac 19+\frac 49+\frac{140}{3}}\\~~=\sqrt{\frac{1+4+420}{9}}\\~~=\sqrt{\frac{425}{9}}\\~~=\sqrt{\frac{25 \times 17}{9}}\\~~=\frac{5\sqrt{17}}{3}$

$r_2-r_1=\frac{5\sqrt{17}}{3}-\sqrt{17}=\frac{2\sqrt{17}}{3}\rightarrow(4)$

Distance between the centres $~(AB)~$ of two circles are

$=\sqrt{(1-\frac 13)^2+(2+\frac 23)^2}\\~~=\sqrt{(2/3)^2+(8/3)^2}\\~~=\frac 13\sqrt{2^2+8^2}\\~~=\frac 13 \sqrt{4+64}\\~~=\frac 13\sqrt{68}\\~~=\frac{2\sqrt{17}}{3}\rightarrow(5)$