Class 11 Maths NCERT Solutions Chapter 10 Ex 10.1 – Conic Sections

Using (x-h)^2+(y-k)^2=r^2 with h=0,\ k=2,\ r=2:

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

Using (x-h)^2+(y-k)^2=r^2 with h=-2,\ k=3,\ r=4:

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

Using (x-h)^2+(y-k)^2=r^2 with h=\dfrac12,\ k=\dfrac14,\ r=\dfrac1{12}:

\left(x-\dfrac12\right)^2+\left(y-\dfrac14\right)^2=\dfrac{1}{144}

Using (x-h)^2+(y-k)^2=r^2 with h=1,\ k=1,\ r=\sqrt2:

(x-1)^2+(y-1)^2=(\sqrt2)^2

Using (x-h)^2+(y-k)^2=r^2 with h=-a,\ k=-b,\ r=\sqrt{a^2-b^2}:

(x+a)^2+(y+b)^2=a^2-b^2

Comparing with (x-h)^2+(y-k)^2=r^2, i.e. writing the equation as \{x-(-5)\}^2+(y-3)^2=6^2:

Grouping the x-terms and y-terms:

(x^2-4x)+(y^2-8y)=45

Completing the square in each bracket:

(x^2-4x+4)+(y^2-8y+16)=45+4+16

(x-2)^2+(y-4)^2=65

Grouping the x-terms and y-terms:

(x^2-8x)+(y^2+10y)=12

Completing the square in each bracket:

(x^2-8x+16)+(y^2+10y+25)=12+16+25

(x-4)^2+(y+5)^2=53

Dividing throughout by 2 to make the coefficients of x^2 and y^2 equal to 1:

x^2+y^2-\dfrac{x}{2}=0

Completing the square in x:

\left(x^2-\dfrac{x}{2}+\dfrac{1}{16}\right)+y^2=\dfrac{1}{16}

\left(x-\dfrac14\right)^2+y^2=\left(\dfrac14\right)^2

Let the equation of the circle be (x-h)^2+(y-k)^2=r^2. Since the circle passes through (4,1) and (6,5), both points give the same r^2:

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

Expanding both sides and simplifying (the squared terms in h and k cancel):

-8h+16-2k+1=-12h+36-10k+25 \ \Rightarrow\ 4h+8k=44 \ \Rightarrow\ h+2k=11 \qquad \ldots(1)

Since the centre lies on 4x+y=16:

4h+k=16 \qquad \ldots(2)

Solving (1) and (2): from (1), h=11-2k. Substituting into (2):

4(11-2k)+k=16 \ \Rightarrow\ 44-8k+k=16 \ \Rightarrow\ -7k=-28 \ \Rightarrow\ k=4

Then h=11-2(4)=3. So the centre is (3,4). Using point (4,1) to find r^2:

r^2=(4-3)^2+(1-4)^2=1+9=10

Let the centre be (h,k). Since the circle passes through (2,3) and (–1,1), both give the same r^2:

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

Expanding and simplifying:

-4h+4-6k+9=2h+1-2k+1 \ \Rightarrow\ -6h-4k+11=0 \ \Rightarrow\ 6h+4k=11 \qquad\ldots(1)

Since the centre lies on x-3y-11=0:

h-3k=11 \ \Rightarrow\ h=11+3k \qquad\ldots(2)

Substituting (2) into (1):

6(11+3k)+4k=11 \ \Rightarrow\ 66+18k+4k=11 \ \Rightarrow\ 22k=-55 \ \Rightarrow\ k=-\dfrac52

Then h=11+3\left(-\dfrac52\right)=\dfrac{7}{2}. So the centre is \left(\dfrac72,-\dfrac52\right). Using point (2,3) to find r^2:

r^2=\left(2-\dfrac72\right)^2+\left(3+\dfrac52\right)^2=\left(-\dfrac32\right)^2+\left(\dfrac{11}{2}\right)^2=\dfrac{9}{4}+\dfrac{121}{4}=\dfrac{130}{4}=\dfrac{65}{2}

Since the centre lies on the x-axis, it has the form (h,0). Using the point (2,3) with radius 5:

(2-h)^2+(3-0)^2=25

(2-h)^2=25-9=16 \ \Rightarrow\ 2-h=\pm4

h=-2 \text{ or } h=6

Making intercepts a and b on the axes means the circle also passes through (a,0) and (0,b), in addition to the origin.

Let the general equation of the circle be x^2+y^2+2gx+2fy+c=0.

Substituting (0,0): c=0.

Substituting (a,0): a^2+2ga=0 \ \Rightarrow\ g=-\dfrac{a}{2} (since a\ne0).

Substituting (0,b): b^2+2fb=0 \ \Rightarrow\ f=-\dfrac{b}{2} (since b\ne0).

Substituting g,f,c back into the general equation:

The radius is the distance from the centre (2,2) to the point (4,5):

r^2=(4-2)^2+(5-2)^2=4+9=13

Using (x-h)^2+(y-k)^2=r^2 with h=2,\ k=2:

The given circle has centre (0,0) and radius 5 (since r^2=25). The squared distance from the centre to the point (–2.5, 3.5) is:

(-2.5)^2+(3.5)^2=6.25+12.25=18.5

Since 18.5<25, the point is closer to the centre than the radius.