Circle |Part-1|S N Dey Examhoop

Today we are going to discuss the Short Answer Type Questions of Circle of S N Dey mathematics Class 11.

Circle Part 1, S N Dey mathematics, Class 11

S N Dey Math Solutions related to Circle (Ex -3)

$1.~$ Examine whether the equation $~x^2+y^2-x-4y+7=0~$ represents a circle.

Solution.

$x^2+y^2-x-4y+7=0 \\ \text{or,}~~ x^2-2 \cdot x \cdot \frac 12+(1/2)^2+y^2-2 \cdot y \cdot 2+2^2=-7+2^2+(1/2)^2 \\ \text{or,}~~\left(x-\frac 12\right)^2 +(y-2)^2=-7+4+\frac 14 \\ \text{or,}~~\left(x-\frac 12\right)^2 +(y-2)^2=\frac 14-3 \\ \text{or,}~~ \left(x-\frac 12\right)^2 +(y-2)^2=\frac{1-12}{4} \\ \text{or,}~~\left(x-\frac 12\right)^2 +(y-2)^2=\left(\frac{\sqrt{-11}}{2}\right)^2 \rightarrow(1)$

Hence, by $~(1),~$ we notice that the radius of the circle being a imaginary number and so, the given equation does not represent a real circle.

$2.~$ Find the equation of the circle passing through $~(6,-5)~$ and having centre at $~(3,-1).$

Solution.

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

$x^2+y^2+2gx+2fy+c=0 \rightarrow(1),$ where $~(-g,-f)~$ is the centre of the circle given by $~(1).~$

$\text{Now,}~ -g=3,~-f=-1 \\ \text{or,}~~g=-3,~f=1 \rightarrow(2)$

Since the circle passing through $~(6,-5)~$ , we get by $~(1)~$ and $~(2),~$

$6^2+(-5)^2+2 \times (-3) \times 6+2 \times 1 \times (-5)+c=0 \\ \text{or,}~~36+25-36-10+c=0 \\ \text{or,}~~ 15+c=0 \\ \text{or,}~~c=-15\rightarrow(3)$

Hence, by $~(1),~(2),~(3)~$ we get, the equation of the circle

$~x^2+y^2+2 \times (-3) \times x+ 2 \times 1 \times y-15=0 \\ \text{or,}~~ x^2+y^2-6x+2y-15=0.$

$3.~$ Find the centre and radius of each of the following circles :

$~(i)~~4x^2+4y^2=25.$

Solution.

We know that the equation of the circle having centre at $~(\alpha,\beta)~$ and radius $~a~$ unit is

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

Now, $~4x^2+4y^2=25~$ can be written as :

$~x^2+y^2=\frac{25}{4} \\ \text{or,}~~ (x-0)^2+(y-0)^2=(5/2)^2\rightarrow(2)$

Hence, from $~(1)~$ and $~(2)~$ , we get, the centre of the given circle is $~(0,0)~$ and the radius is $~(5/2)$ unit.

$(ii)~~x^2+y^2-3x+2y-19=0$

Solution.

$x^2+y^2-3x+2y-19=0 \\ \text{or,}~~ x^2-2 \cdot x \cdot \frac 32+(3/2)^2+y^2+2 \cdot y \cdot 1+1^2\\=19+(3/2)^2+1^2 \\ \text{or,}~~ \left(x-\frac 32\right)^2 +(y+1)^2=20+\frac 94 \\ \text{or,}~~ \left(x-\frac 32\right)^2 +(y+1)^2=\frac{89}{4} \\ \text{or,}~~ \left(x-\frac 32\right)^2 +[y-(-1)]^2 =\left(\frac{\sqrt{89}}{2}\right)^2 \rightarrow(1)$

Hence, from $~(1),~$ we get, the centre of the given circle is $~(3/2,-1)~$ and the radius is $~\left(\frac{\sqrt{89}}{2}\right)$ unit.

$(iii)~~3(x^2+y^2)=5x+6y-4$

Solution.

$~3(x^2+y^2)=5x+6y-4 \\ \text{or,}~~ x^2+y^2=\frac 53x+2y-\frac 43 \\ \text{or,}~~x^2-\frac 53 x+y^2-2y=-\frac 43 \\ \text{or,}~~ x^2-2 \cdot x \cdot (5/6) +(5/6)^2+y^2-2 \cdot y \cdot 1+1^2 \\ =-\frac 43+(5/6)^2+1^2 \\ \text{or,}~~ (x-5/6)^2+(y-1)^2=-\frac 13+\frac{25}{36} \\~~ \\ \text{or,}~~ (x-5/6)^2+(y-1)^2 =\frac{-12+25}{36} \\ \text{or,}~~ (x-5/6)^2+(y-1)^2 =\frac{13}{36}=\left(\frac{\sqrt{13}}{6}\right)^2 \rightarrow(1)$

Hence, from $~(1),~$ we get, the centre of the given circle is $~(5/6,1)~$ and the radius is $~\left(\frac{\sqrt{13}}{6}\right)$ unit.

$(iv)~~(x-a)^2+(y+b)(y-b)=0$

Solution.

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

Hence, from $~(1),~$ we get, the centre of the given circle is $~(a,0)~$ and the radius is $~b~$ unit.

4. Under what conditions $~ax^2+2hxy+by^2+2gx+2fy+c=0~$ will be the equation of a circle? In that case what will be the co-ordinates of the centre ?

Solution.

$ax^2+2hxy+by^2+2gx+2fy+c=0~$ will represent an equation of a circle if

$(i)~ a=b(\neq 0), ~(ii)~ h=0.$

For $a=b(\neq 0),~h=0~$ the given equation can be written as follows :

$ax^2+ay^2+2gx+2fy+c=0 \\ \text{or,}~~x^2+y^2+ \frac{2gx}{a}+\frac{2fy}{a}+c/a=0 \rightarrow(1)$

Comparing the equation $~(1),~$ with the general / standard form of circle, we can say that the centre of the circle is : $~(-g/a,-f/a).$

$5.~~$ Find the radius of the circle which passes through the origin and the points $~(a,0)~$ and $~(0,b).$

Solution.

The general equation of the circle is given by :

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

Since the circle which passes through the origin, we get $~(1)~$ , $~c=0.$

So, the equation $~(1)~$ can be rewritten as

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

Since the circle passes through the point $~(a,0)~$ , we get by $~(2),$

$~a^2+0+2ga+0=0 \\ \Rightarrow a(a+2g)=0 \\ \therefore g=-a/2.$

Again, since the circle passes through the point $~(0,b),~$ we get by $~(2),$

$0+b^2+0+2fb=0 \\ \Rightarrow b(b+2f) =0 \\ \therefore f=-b/2.$

Hence, the radius of the circle as represented by $~(2),~$ is given by :

$~\sqrt{g^2+f^2}=\sqrt{(-a/2)^2+(-b/2)^2}=\frac 12 \sqrt{a^2+b^2}~~\text{unit (ans.)}$

$6.~$ Find the equation of the circle for which the line segment joining the points $~(i)~(2a,0)~$ and $~(0,-2a)~~(ii) ~~(3,7)~$ and $~(9,1)~$ is a diameter.

Solution.

According to the problem, The mid-point of the line segment joining the points $~(2a,0)~$ and $~(0,-2a)~$ will be the centre of the circle and the centre of the circle is :

$~(\alpha,\beta)=\left(\frac{2a+0}{2},\frac{0-2a}{2}\right)=(a,-a).$

The diameter of the circle is the length of the line segment joining the points $~~(2a,0)~$ and $~~(0,-2a)~$ and is given by :

$\sqrt{(2a-0)^2+(0+2a)^2}\\~~~=\sqrt{4a^2+4a^2}\\~~~=\sqrt{2 \times 4a^2}\\~~~=2\sqrt{2}a$

and so, the radius of the circle is :

$r=\frac{2\sqrt{2}a}{2}=\sqrt{2}a~\text{unit.}$

Hence, the required equation of the circle is :

$(x-\alpha)^2+(y-\beta)^2=(\sqrt{2}a)^2 \\ \text{or,}~~ (x-a)^2+(y+a)^2=2a^2 \\ \text{or,}~~ x^2-2ax+a^2+y^2+2ay+a^2=2a^2 \\ \text{or,}~~ x^2+y^2-2ax+2ay=0.$

2nd Part :

Solution.

According to the problem, The mid-point of the line segment joining the points $~~(3,7)~$ and $~~(9,1)~$ will be the centre of the circle and the centre of the circle is :

$~(\alpha,\beta)=\left(\frac{3+9}{2},\frac{7+1}{2}\right)=(6,4).$

The diameter of the circle is the length of the line segment joining the points $~~(3,7)~$ and $~~(9,1)~$ and is given by :

$\sqrt{(3-9)^2+(7-1)^2}=\sqrt{(-6)^2+6^2}=\sqrt{2 \times 36}=6\sqrt{2}$

and so, the radius of the circle is : $r=\frac{6\sqrt{2}}{2}=3\sqrt{2} ~~\text{unit.}$

Hence, the required equation of the circle is :

$(x-\alpha)^2+(y-\beta)^2=(3\sqrt{2})^2 \\ \text{or,}~~(x-6)^2+(y-4)^2=18 \\ \text{or,}~~x^2-2 \cdot x \cdot 6+6^2+y^2-2 \cdot y \cdot 4+4^2=18 \\ \text{or,}~~x^2+y^2-12x-8y+36+16-18=0 \\ \text{or,}~~ x^2+y^2-12x-8y+34=0~~\text{(ans.)}$

$7.~$ Find the position of the point $~(-3,-2)~$ with respect to the circle whose equation is $~ x^2+y^2-3x+2y-19=0.$

Solution.

Comparing the given equation of the circle with the general equation / standard form of circle i.e., with $~ x^2+y^2+2gx+2fy+c=0~$ , we get,

$2g=-3 \Rightarrow g=-\frac 32, \\ ~~ 2f=2 \Rightarrow f=\frac 22=1, \\~~ c=-19.$

Hence, the radius of the circle is :

$~\sqrt{g^2+f^2-c}\\=\sqrt{(-3/2)^2+(-1)^2+19}\\=\sqrt{\frac{9}{4}+1+19}\\=\sqrt{\frac{9}{4}+20}\\=\sqrt{\frac{89}{4}} \rightarrow(1)$

Again, the centre of the circle is given by : $(-g,-f)=(3/2,-1).$

Now, the distance of the point $~(-3,-2)~$ from the centre of the circle i.e., $~(3/2,1)~$ is :

$\sqrt{(3/2+3)^2+(-1+2)^2}\\=\sqrt{\left(\frac{9}{2}\right)^2+(1)^2}\\=\sqrt{\frac{81}{4}+1}\\=\sqrt{\frac{81+4}{4}}\\=\sqrt{\frac{85}{4}}\rightarrow(2)$

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

$\sqrt{\frac{85}{4}} < \sqrt{\frac{89}{4}}$ and so, the point $~(-3,-2)~$ lies inside the given circle.

$9.$ The straight line $~3x-4y+7=0~$ is a tangent to the circle $~x^2+y^2+4x+2y+4=0~$ at P ; find the equation of its normal at the same point.

Solution.

Given that straight line $~3x - 4y + 7 = 0~$ is tangent to the circle $~x^2+y^2+4x+2y+4=0~$ .

Now, General equation of circle $~x^2 + y^2 + 2gx + 2fy + c = 0,~$ where center of circle is : $~ ( -g, -f)~$ and radius of circle is : $~\sqrt{g^2+f^2-c}.$

Comparing the given equation to general equation we get,

$~2g=4 \Rightarrow g=\frac 42=2,\~~2f=2 \Rightarrow f=\frac 22=1.$

So, center of circle $: (-g,-f)= (-2,-1)$ .

Now, the equation of the straight line which is perpendicular to the given straight line is : $~4x+3y+k=0~$ which passes through the point $~(-2,-1)~$ and so,

$~4 \times (-2)+3 \times (-1)+k=0\\~~ \Rightarrow -8-3+k=0 \\ ~~\Rightarrow k=8+3=11 \rightarrow(1).$

Hence, the equation of its normal at the same point is given by :

$~4x+3y+11=0 ~[\text{By (1)}].$

$10.~~$ The length of diameter of the circle $~x^2+y^2+4x-7y-k=0~$ is $~9;~$ find $~k.$

Solution.

General equation of circle $~x^2 + y^2 + 2gx + 2fy + c = 0,~$ where center of circle is : $~ ( -g, -f)~$ and radius of circle is : $~\sqrt{g^2+f^2-c}.$

Comparing the given equation to general equation we get,

$2g=4 \Rightarrow g=\frac 42=2, \\~~ 2f=-7 \Rightarrow f=\frac{-7}{2},\\~~ c=-k.$

So, center of circle $: (-g,-f)= (-2,7/2)$ .

The radius of the circle : $\sqrt{2^2+(-7/2)^2+k}.$

So, according to the given problem,

$2\sqrt{2^2+(-7/2)^2+k}=9 \\ \text{or,}~~ 4+\frac{49}{4}+k=(9/2)^2 \\ \text{or,}~~ k=\frac{81}{4}-4-\frac{49}{4} \\ \text{or,}~~ k=\frac{81-49}{4}-4 \\ \text{or,}~~ k=\frac{32}{4}-4 \\ \text{or,}~~ k=8-4 \\ \text{or,}~~k=4.$

Post Views: 153

S N Dey Math Solutions related to Circle (Ex -3)

Leave a Comment Cancel reply