Question

question 46

46. Equation of circles touching x-axis at the origin and the line 4x-3y+24=0 are
1) x2+y2-6y=0 , x2+y2+24y=0,
2) x2-y2+2y=0, x2+y2+24=0,
3) x2+y2+18x=0, x2+y2-8x=0,
4) x2+y2+4x=0, x2+y2-16x=0,

Answer 1

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Conic Sections, Class 11 Mathematics R.D Sharma Solutions

CIRCLESPage 24.16 Ex 24.1

Q1.

Answer :

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be x-h2+y-k2=a2.

(i) Here, h = −2, k = 3 and a = 4

∴ Required equation of the circle:
x+22+y-32=42
⇒x+22+y-32=16

(ii) Here, h = a, k = b and radius = a2+b2

∴ Required equation of the circle:
x-a2+y-b2=a2+b2
⇒x2+y2-2ax-2by=0

(iii) Here, h = 0, k = −1 and radius = 1

∴ Required equation of the circle:
x-02+y+12=12
⇒x2+y2+2y=0

(iv) Here, h = acosα, k = asinα and radius = a

∴ Required equation of the circle:
x-acosα2+y-asinα2=a2
⇒x2+a2cos2α-2axcosα+y2+a2sin2α-2aysinα=a2⇒x2+a2sin2α+cos2α-2axcosα+y2-2aysinα=a2⇒x2+a2-2axcosα+y2-2aysinα=a2⇒x2+y2-2axcosα-2aysinα=0

(v) Here, h = a, k = a and radius = 2a

∴ Required equation of the circle:
x-a2+y-a2=2a2
⇒x2+a2-2ax+y2+a2-2ay=2a2⇒x2+y2-2ay-2ax=0

Q2.

Answer :

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be x-h2+y-k2=a2.

(i) Given:
(x − 1)2 + y2 = 4

Here, h = 1, k = 0 and a = 2

Thus, the centre is (1, 0) and the radius is 2.

(ii) Given:
(x + 5)2 + (y + 1)2 = 9

Here, h = −5, k = −1 and radius = 3

Thus, the centre is (−5, −1) and the radius is 3.

(iii) Given:
x2+y2-4x+6y=5

The given equation can be rewritten as follows:
x-22+y+32-4-9=5
⇒x-22+y+32=18

Thus, the centre is (2, −3).
And, radius = 18=32

(iv) Given:
x2+y2-x+2y-3=0

The given equation can be rewritten as follows:
x-122+y+12-14-1-3=0
⇒x-122+y+12=174

Thus, the centre is 12,-1 and and the radius is 172.

Q3.

Answer :

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be x-h2+y-k2=a2.

Given:
h = 1, k = 2

∴ Equation of the circle = x-12+y-22=a2 …(1)

Also, equation (1) passes through (4, 6).

∴ 4-12+6-22=a2
⇒9+16=a2⇒a=5 ∵a>0

Substituting the value of a in equation (1):

x-12+y-22=25

⇒x2+1-2x+y2+4-4y=25⇒x2-2x+y2-4y=20⇒x2+y2-2x-4y-20=0

Thus, the required equation of the circle is x2+y2-2x-4y-20=0.

Q4.

Answer :

The point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 is (0, 0).

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be x-h2+y-k2=a2.

The point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0 is (−2,


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