NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 - Conic Section

Question:2 Find the equation of the circle with

centre (–2,3) and radius 4

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(-2,3)$

AND $r=4$

So the equation of the circle is:

, $(x-(-2))^2+(y-3)^2=4^2$

$x^2+4x+4+y^2-6y+9=16$

$x^2+y^2+4x-6y-3=0$

Question:3 Find the equation of the circle with

centre $\left(\frac{1}{2},\frac{1}{4} \right )$ and radius $\frac{1}{12}$

Answer:

As we know,

The equation of the circle with center ( h, k) and radius r is give by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=\left ( \frac{1}{2},\frac{1}{4} \right )$

AND

$r=\frac{1}{12}$

So the equation of circle is:

, $\left ( x-\frac{1}{2}\right )^2+\left ( y-\frac{1}{4}\right )^2=\left ( \frac{1}{12}\right )^2$

$x^2-x+\frac{1}{4}+y^2-\frac{1}{2}y+\frac{1}{16}=\frac{1}{144}$

$x^2+y^2-x-\frac{1}{2}y-\frac{11}{36}=0$

$36x^2+36y^2-36x-18y-11=0$

Question:4 Find the equation of the circle with

centre (1,1) and radius $\sqrt2$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(1,1)$

AND $r=\sqrt{2}$

So the equation of the circle is:

, $(x-1)^2+(y-1)^2=(\sqrt{2})^2$

$x^2-2x+1+y^2-2y+1=2$

$x^2+y^2-2x-2y=0$

Question:5 Find the equation of the circle with

centre $(-a,-b)$ and radius $\sqrt{a^2 - b^2}$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(-a,-b)$

AND $r=\sqrt{a^2-b^2}$

So the equation of the circle is:

, $(x-(-a))^2+(y-(-b))^2=(\sqrt{a^2-b^2})^2$

$x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2$

$x^2+y^2+2ax+2by+2b^2=0$

Question:6 Find the centre and radius of the circles.

$(x+5)^2 + (y-3)^2 = 36$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$(x+5)^2 + (y-3)^2 = 36$

Can also be written in the form

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

So, from comparing, we can see that

$r=6$

Hence the Radius of the circle is 6.

Question:7 Find the centre and radius of the circles.

$x^2 + y^2 -4x - 8y - 45 = 0$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$x^2 + y^2 -4x - 8y - 45 = 0$

Can also be written in the form

$(x-2)^2 + (y-4)^2 =(\sqrt{65})^2$

So, from comparing, we can see that

$r=\sqrt{65}$

Hence the Radius of the circle is $\sqrt{65}$.

Question:8 Find the centre and radius of the circles.

$x^2 + y^2 -8x +10y -12 = 0$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$x^2 + y^2 -8x +10y -12 = 0$

Can also be written in the form

$(x-4)^2 + (y-(-5))^2 = (\sqrt{53})^2$

So, from comparing, we can see that

$r=\sqrt{53}$

Hence the radius of the circle is $\sqrt{53}$.

Question:9 Find the centre and radius of the circles.

$2x^2 + 2y^2 - x = 0$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$2x^2 + 2y^2 - x = 0$

Can also be written in the form

$\left ( x-\frac{1}{4}\right )^2 + \left ( y-0 \right )^2 = \left ( \frac{1}{4} \right )^2$

So, from comparing, we can see that

$r=\frac{1}{4}$

Hence Center of the circle is the $\left ( \frac{1}{4},0\right )$Radius of the circle is $\frac{1}{4}$.

Question:10 Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line $4x + y = 16$.

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (4,1) and (6,5)

$(4-h)^2+(1-k)^2=r^2$

$(6-h)^2+(5-k)^2=r^2$

Here,

$(4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2$

$(4-h)^2-(6-h)^2+(1-k)^2-(5-k)^2=0$

$(-2)(10-2h)+(-4)(6-2k)=0$

$-20+4h-24+8k=0$

$4h+8k=44$

Now, Condition 2:centre is on the line $4x + y = 16$.

$4h+k=16$

From condition 1 and condition 2

$h=3,\:k=4$

Now lets substitute this value of h and k in condition 1 to find out r

$(4-3)^2+(1-4)^2=r^2$

$1+9=r^2$

$r=\sqrt{10}$

So now, the Final Equation of the circle is

$(x-3)^2+(y-4)^2=(\sqrt{10})^2$

$x^2-6x+9+y^2-8y+16=10$

$x^2+y^2-6x-8y+15=0$

Question:11 Find the equation of the circle passing through the points (2,3) and (–1,1) and hose centre is on the line $x - 3y - 11 = 0$.

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (2,3) and (–1,1)

$(2-h)^2+(3-k)^2=r^2$

$(-1-h)^2+(1-k)^2=r^2$

Here,

$(2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$

$(2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0$

$(3)(1-2h)+(2)(4-2k)=0$

$3-6h+8-4k=0$

$6h+4k=11$

Now, Condition 2: centre is on the line.$x - 3y - 11 = 0$

$h-3k=11$

From condition 1 and condition 2

$h=\frac{7}{2},\:k=\frac{-5}{2}$

Now let's substitute this value of h and k in condition 1 to find out $r$

$\left ( 2-\frac{7}{2}\right )^2+\left (3+\frac{5}{2}\right )^2=r^2$

$\frac{9}{4}+\frac{121}{4}=r^2$

$r^2=\frac{130}{4}$

So now, the Final Equation of the circle is

$\left(x-\frac{7}{2}\right )^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4}$

$x^2-7x+\frac{49}{4}+y^2+5y+\frac{25}{4}=\frac{130}{4}$

$x^2+y^2-7x+5y-\frac{56}{4}=0$

$x^2+y^2-7x+5y-14=0$

Question:12 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So let the circle be,

$(x-h)^2+(y-k)^2=r^2$

Since it's radius is 5 and its centre lies on x-axis,

$(x-h)^2+(y-0)^2=5^2$

And Since it passes through the point (2,3).

$(2-h)^2+(3-0)^2=5^2$

$(2-h)^2=25-9$

$(2-h)^2=16$

$(2-h)=4\:or\:(2-h)=-4$

$h=-2\: or\;6$

When $h=-2\:$ ,The equation of the circle is :

$(x-(-2))^2+(y-0)^2=5^2$

$x^2+4x+4+y^2=25$

$x^2+y^2+4x-21=0$

When $h=6$ The equation of the circle is :

$(x-6)^2+(y-0)^2=5^2$

$x^2-12x+36+y^2=25$

$x^2+y^2-12x+11=0$

Question:13 Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Answer:

Let the equation of circle be,

$(x-h)^2+(y-k)^2=r^2$

Now since this circle passes through (0,0)

$(0-h)^2+(0-k)^2=r^2$

$h^2+k^2=r^2$

Now, this circle makes an intercept of a and b on the coordinate axes.it means circle passes through the point (a,0) and (0,b)

So,

$(a-h)^2+(0-k)^2=r^2$

$a^2-2ah+h^2+k^2=r^2$

$a^2-2ah=0$

$a(a-2h)=0$

$a=0\:or\:a-2h=0$

Since $a\neq0\:so\:a-2h=0$

$h=\frac{a}{2}$

Similarly,

$(0-h)^2+(b-k)^2=r^2$

$h^2+b^2-2bk+k^2=r^2$

$b^2-2bk=0$

$b(b-2k)=0$

Since b is not equal to zero.

$k=\frac{b}{2}$

So Final equation of the Circle ;

$\left ( x-\frac{a}{2} \right )^2+\left ( y-\frac{b}{}2 \right )^2=\left ( \frac{a}{2} \right )^2+\left ( \frac{b}{2} \right )^2$

$x^2-ax+\frac{a^2}{4}+y^2-bx+\frac{b^2}{4}=\frac{a^2}{4}+\frac{b^2}{4}$

$x^2+y^2-ax-bx=0$

Question:14 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Answer:

Let the equation of circle be :

$(x-h)^2+(y-k)^2=r^2$

Now, since the centre of the circle is (2,2), our equation becomes

$(x-2)^2+(y-2)^2=r^2$

Now, Since this passes through the point (4,5)

$(4-2)^2+(5-2)^2=r^2$

$4+9=r^2$

$r^2=13$

Hence The Final equation of the circle becomes

$(x-2)^2+(y-2)^2=13$

$x^2-4x+4+y^2-4y+4=13$

$x^2+y^2-4x-4y-5=0$

Question:15 Does the point (–2.5, 3.5) lie inside, outside or on the circle $x^2 + y^2 = 25$?

Answer:

Given, a circle

$x^2 + y^2 = 25$

As we can see center of the circle is ( 0,0)

Now the distance between (0,0) and (–2.5, 3.5) is

$d=\sqrt{(-2.5-0)^2+(3.5-0)^2}$

$d=\sqrt{6.25+12.25}$

$d=\sqrt{18.5}\approx 4.3$$d=\sqrt{18.5}\approx 4.3<5$

Since distance between the given point and center of the circle is less than radius of the circle, the point lie inside the circle.