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) = (x_{1}, y_{1})

S_{11} = 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:

S_{11} = (1)² + (5)² – 2(-1) – 4(5) + 3 = 7

S_{11} > 0 [∴ P is outside the circle]

iii) P (4, 2) and S ≡ 2x² + 2y² – 5x – 4y – 3 = 0

Solution:

S_{11} = 2(4)² + 2(2)² – 5(4) – 4(2) – 3 = 9

S_{11} > 0 (P lies outside the circle)

iv) P(2, -1) and S ≡ x² + y² – 2x – 4y + 3 = 0

Solution:

S_{11} = (2)² + (-1)² – 2(2) – 4 (-1) + 3 = 8

S_{11} > 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:

S_{11} = 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 = S_{11}

= 1+1+6 + 4-12 = 0

iii) P = (2, 3) and S ≡ x² + y² – 2x + 8y – 23

Power of the point = S_{11}

= 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 (x_{1}, y_{1}) is any point on the locus and PT_{1} PT_{2} are the tangents from P to the two circles.

Given condition is PT_{1} = PT_{2} ⇒ P_{1}T_{1}² = PT_{2}²

x²_{1} + y²_{1} + 8x_{1} + 12y_{1} + 15

= x²_{1} + y²_{1} – 4x_{1} – 6y_{1} – 12

12x_{1} + 18y_{1} + 27 = 0

(or) 4x_{1} + 6y_{1} + 9 = 0

Locus of P(x_{1}, y_{1}) is 4x + 6y + 9 = 0