Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(b)

I.

Question 1.
Locate the position of the point P with respect to the circle S = 0 when
i) P(3, 4) and S ≡ x² + y² – 4x – 6y – 12 = 0
Solution:
S ≡ x² + y² – 4x – 6y – 12
P(3, 4) = (x1, y1)
S11 = 3² + 4² – 4.3 – 6.4 – 12
= 9 + 16 – 12 – 24 – 12
= – 23 < 0
P (3, 4) lies inside the circle

ii) P(1, 5) and S ≡ x² + y² – 2x – 4y + 3 = 0
Solution:
S11 = (1)² + (5)² – 2(-1) – 4(5) + 3 = 7
S11 > 0 [∴ P is outside the circle]

iii) P (4, 2) and S ≡ 2x² + 2y² – 5x – 4y – 3 = 0
Solution:
S11 = 2(4)² + 2(2)² – 5(4) – 4(2) – 3 = 9
S11 > 0 (P lies outside the circle)

iv) P(2, -1) and S ≡ x² + y² – 2x – 4y + 3 = 0
Solution:
S11 = (2)² + (-1)² – 2(2) – 4 (-1) + 3 = 8
S11 > 0 [P is outside the circle]

Question 2.
Find the power of the point P with respect to the circle S = 0 when
i) P = (5, -6), and S ≡ x² + y² + 8x + 12y + 15
Solution:
S11 = power of the point
= 25 + 36 + 40 – 72 + 15
= 116 – 72 = 44

ii) P = (-1, 1) and S ≡ x² + y² -6x + 4y – 12
Power of the point = S11
= 1+1+6 + 4-12 = 0

iii) P = (2, 3) and S ≡ x² + y² – 2x + 8y – 23
Power of the point = S11
= 4 + 9-4 + 24-23 = 10

iv) P = (2, 4) and S ≡ x² + y² – 4x – 6y – 12
Power of the point = 4 + 16 – 8 – 24 – 12
= -24. Question 3.
Find the length of tangent from P to the circle S = 0 when
i) P = (-2, 5) and S = x² + y² – 25
Solution:
Length of tangent = $$\sqrt{s_{11}}$$
= $$\sqrt{(-2)^{2}+(5)^{2}-25}$$ = 2 units

ii) P = (0, 0), S = x² + y² – 14x + 2y + 25
Solution:
Length of the tangent = $$\sqrt{s_{11}}$$
= $$\sqrt{0+0-0+0+25}$$ = 5 units

iii) P = (2, 5) and S ≡ x² + y² – 5x + 4y – 5
Solution:
Length of the tangent = $$\sqrt{s_{11}}$$
= $$\sqrt{4+25-10+20-5}$$
= $$\sqrt{34}$$ units

II.

Question 1.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is 1 then find k.
Solution:
Length of tangent
$$\sqrt{s_{11}}=\sqrt{(5)^{2}+(4)^{2}+8k}$$
But length of tangent = 1
∴ 1 = $$\sqrt{25+16+8k}$$
Squaring both sides we get 1 = 41 + 8k
k = – 5 units.

Question 2.
If the length of the tangent from (2, 5) to the circle x² + y² – 5x + 4y + k = 0 is √37 then find k.
Solution:
Length of tangent = $$\sqrt{s_{11}}$$
= $$\sqrt{(2)^2+(5)^2-5 \times 2+4 \times 5+k}$$
= 37 = 39 + k
k = -2 units. III.

Question 1.
If a point P is moving such that the lengths of tangents drawn from P to the circles x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0 are in the ratio 2 : 3, then find the equation of the locus of P.
Solution:
P(x, y) is any point on the locus
S ≡ x² + y² – 4x – 6y – 12 Locus of P is
5x² + 5y² – 60x – 126y – 212 = 0 Question 2.
If a point P is moving such that the lengths of the tangents drawn from P to the circles x² + y² + 8x + 12y + 15 = 0 and x² + y² – 4x – 6y – 12 = 0 are equal then find the equation of the locus of P. Solution:
Equations of the circles are
S ≡ x² + y² + 8x + 12y + 15 = 0
S¹ ≡ x² + y² – 4x – 6y – 12 = 0
P (x1, y1) is any point on the locus and PT1 PT2 are the tangents from P to the two circles.
Given condition is PT1 = PT2 ⇒ P1T1² = PT2²
1 + y²1 + 8x1 + 12y1 + 15
= x²1 + y²1 – 4x1 – 6y1 – 12
12x1 + 18y1 + 27 = 0
(or) 4x1 + 6y1 + 9 = 0
Locus of P(x1, y1) is 4x + 6y + 9 = 0