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

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

I.

Question 1.

Find the condition that the tangents drawn from (0, 0) to S ≡ x² + y² + 2gx + 2fy + c = 0 be perpendicular to each other.

Solution:

If θ is the angle between the pair of tangents from P_{1} then tan \(\frac{\theta}{2}=\frac{r}{\sqrt{s_{11}}}\)

g² + f² – c = c

g² + f² = 2c

This is the required condition.

Question 2.

Find the chord of contact of (0, 5) with respect to the circle x² + y² – 5x + 4y – 2 = 0

Solution:

Equation of the chord of contact is S_{1} = 0

i.e., xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

Equation of the circle is

S = x² + y² – 5x + 4y – 2 = 0

Equation of the chord of contact is

x. 0 + y. 5 – \(\frac{5}{2}\) (x + 0) + 2(y + 5) – 2 = 0

Multiplying with 2

10y – 5x + 4y + 20 – 4 = 0 – 5x + 14y +16 = 0

or 5x – 14y – 16 = 0

Question 3.

Find the chord of contact of (1, 1) to the circle x² + y² = 9.

Solution:

Equation of the circle is x² + y² = 9

Equation of the chord of contact is

x.1 + y. 1 =9

i.e. x + y = 9

Question 4.

Find the polar of (1, 2) with respect to x² + y² = 7.

Solution:

Polar of P(x_{1}, y_{1}) w.r. to S = 0

xx_{1} + yy_{1} = a²

x + 2y – 7 = 0 is the polar equation.

Question 5.

Find the polar of (3, -1) with respect to 2x² + 2y² = 11.

Solution:

Equation of circle is 2x² + 2y² = 11

x² + y² = \(\frac{11}{2}\)

Equation of polar is xx_{1} + yy_{1} = a²

x(3) + y(-1) = \(\frac{11}{2}\)

Equation-of polar is 6x – 2y -11 = 0

Question 6.

Find the polar of (1, -2) with respect to x² + y² – 10x – 10y + 25 = 0

Solution:

Equation of the circle is

x² + y² – 10x – 10y + 25 = 0

Equation of the polar is S_{1} = 0

Polar of P(1, -2) is

x .1 + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0

⇒ x – 2y – 5x – 5 – 5y + 10 + 25 = 0

⇒ -4x – 7y + 30 = 0

∴ 4x + 7y – 30 = 0

Question 7.

Find the pole of ax + by + c = 0 (c + 0) with respect to x² + y² = r².

Solution:

Let (x_{1}, y_{1}) be pole, the polar equation be

xx_{1} + yy_{1} – r² = 0 ………… (i)

ax + by + c = 0 ………….. (ii)

(i) and (ii) both are same (polar) equations.

Question 8.

Find the pole of 3x + 4y – 45 = 0 with respect to x² + y² – 6x – 8y + 5 = 0.

Solution:

Equation of polar is

xx_{1} + yy_{1} – 3(x + x_{1}) – 4(y + y_{1}) + 5 = 0

x(x_{1} – 3) + y(y_{1} – 4) – 3x_{1} – 4y_{1} + 5 = 0 ……….. (i)

Polar equation is same 3x + 4y – 45 = 0 ……….. (ii)

Comparing (i) and (ii) we get

Pole is (6, 8).

Question 9.

Find the pole of x – 2y + 22 = 0 with respect to x² + y² – 5x + 8y + 6 = 0.

Solution:

Polar equation is

xx_{1} + yy_{1} – \(\frac{5}{2}\)(x + x_{1}) + 4(y + y_{1}) + 6 = 0

(or) x(x_{1} – \(\frac{5}{2}\)) + y(y_{1} + 4) – \(\frac{5}{2}\)

x_{1} + 4y_{1} + 6 = 0 …………. (i)

x – 2y + 22 = 0 ………….. (ii)

(i) and (ii) both are equations of the same line comparing it we get

y_{1} = -3

∴ Pole is (2, -3)

Question 10.

Show that the points (-6, 1) and (2, 3) are conjugate points with respect to the circle x² + y² – 2x + 2y + 1 =0.

Solution:

Polar equation w.r.t. (2, 3) is

2x + 3y – 1(x + 2) + 1(y + 3) + 1 = 0

x + 4y + 2 = 0 ………. (i)

Now (-6, 1) should satisfy equation (i) if it is conjugate point

(-6) + 4(1) + 2 = 0

∴ (-6, 1) and (2, 3) are conjugate w.r.t. circle.

Question 11.

Show that the points (4, 2) (3, -5) are conjugate points with respect to the circle x² + y² – 3x – 5y + 1 = 0.

Solution:

Polar equation w.r.t. (4, 2) is

4x + 2y – \(\frac{3}{2}\)(x + 4) – \(\frac{5}{2}\)(y + 2) + 1 = 0 ………. (i)

Now (3, -5) should satisfy equation (i) if it is conjugate points.

4(3) – 5 × 2 – \(\frac{3}{2}\)(3 + 4) – \(\frac{5}{2}\)(-5 + 2) + 1

12 – 10 – \(\frac{21}{2}\) + \(\frac{15}{2}\) + 1 = 0

Hence given points are conjugate.

Question 12.

Find the value of k if kx + 3y – 1 = 0, 2x + y + 5 = 0 are conjugate lines with respect to the circle x² + y² – 2x – 4y – 4 = 0.

Solution:

Let pole be (x_{1}, y_{1}) polar equation w.r.t.

x² + y² – 2x – 4y – 4 = 0 be

xx_{1} + yy_{1} -1 (x + x_{1}) – 2(y + y_{1}) – 4 = 0

x(x_{1} – 1) + y(y_{1} – 2) – x_{1} – 2y_{1} – 4 = 0 …….. (i)

2x + y + 5 = 0 is polar so comparing with equation (i) we get

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

kx + 3y – 1 = 0 is polar so it should satisfy

(1, 1)

k(-1) + 3(1) – 1 = 0

-k + 2 = 0

k = 2

Question 13.

Find the value of k if x + y – 5 = 0, 2x + ky – 8 = 0 are conjugate with respect to the circle x² + y² – 2x – 2y – 1 = 0.

Solution:

If l_{1}x + m_{1}y + n_{1} = 0; l_{2}x + m_{2}y + n_{2} = 0 are conjugate, S = 0 then,

r²(l_{1}l_{2} + m_{1}m_{2})

= (l_{1}g + m_{1}f – n_{1}) (l_{2}g + m_{2}f – n_{2})

l_{1} = 1, m_{1} = 1, n_{1} = -5

l_{2} = 2, m_{2} = k, n_{2} = – 8

g = -1, f = -1, r² = 3

∴ 3(1.2 + k) = (-1 -1 + 5) (-2 – k + 8)

6k = 18 – 6 = 12 ⇒ k = 2

Question 14.

Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x² + y² = 35.

Solution:

Equation of the circle is x² + y² = 35

Polar of P(1, 3) is x. 1 + y. 3 = 35

x + 3y = 35

P(1, 3) and Q(2, k) are conjugate points The polar of P passes through Q

2 + 3k = 35

3k = 33

k = 11

Question 15.

Find the value of k if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x² + y² – 5x + 8y + 6 = 0.

Solution:

Equation of the circle is x² + y² – 5x + 8y + 6 = 0

Polar of P(4, 2) is

x.4 + y.2 – \(\frac{5}{2}\)(x + 4) + 4(y + 2) + 6 = 0

8x + 4y – 5x – 20 + 8y + 16 + 12 = 0

3x + 12y + 8 = 0

P(4, 2), Q(k, -3) are conjugate points

Polar of P passes through Q

∴ 3k – 36 + 8 = 0

3k = 28 ⇒ k = \(\frac{28}{2}\)

II.

Question 1.

Find the angle between the tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0.

Solution:

Equation of the circle is

S ≡ x² + y² – 6x + 4y – 2 = 0

Angle between the tangent at P = cos^{-1} (\(\frac{7}{8}\))

Question 2.

Find the angle between the pair of tangents dravyn from (1, 3) to the circle x² + y² – 2x + 4y – 11 = 0.

Solution:

Equation of the circle is

S_{1} ≡ x² + y² – 2x + 4y – 11 = 0

r = \(\sqrt{1+4+11}\) = 4

S_{11} =1 +9-2 + 12-11 = 9

Angle between the tangents = cos^{-1} (\(\frac{7}{25}\))

Question 3.

Find the angle between the pair of tangents drawn from (0,0) to the circle x² + y² – 14x + 2y + 25 = 0.

Solution:

Equation of the circle is

x² + y² – 14x + 2y + 25 = 0

r = \(\sqrt{49+1-25}\) = 5

S_{11} = 0 + 0 – 0 + 0 + 25 = 25 ⇒ √S_{11} = 5

tan θ = \(\frac{r}{\sqrt{s_{11}}}\) = \(\frac{5}{5}\) = 1 ⇒ θ = \(\frac{\pi}{4}\)

Angle between the tangents = 2θ = \(\frac{\pi}{2}\)

Question 4.

Find the locus of P if the tangents drawn from P to x² + y²= a² include an angle α.

Solution:

Equation of the circle is x² + y² = a²

r = radius = a

Suppose θ (x_{1}, y_{1}) is any point as the locus

Question 5.

Find the locus of P if the tangents drawn from P to x2 + y2= a2 are perpendicular to each other.

Solution:

Equation of the circle is x² + y² = a²

r = a

Let P(x_{1}, y_{1}) be any point on the locus

S_{11} = x²_{1} + y²_{1} – a²

Squaring and cross-multiplying

a² = x² + y² – a²

x² + y² = 2a²

Locus of P (x_{1}, y_{1}) is x² + y²= 2a²

Question 6.

Find the slope of the polar of (1, 3) with respect to the circle x² + y² – 4x – 4y – 4= 0. Also find the distance from the centre to it.

Solution:

Equation of the circle is x² + y² – 4x – 4y – 4 =0

Polar of P (1, 3) is

x.1 + y.3 – 2(x + 1) – 2(y + 3) – 4 = 0

x + 3y – 2x – 2 – 2y – 6 – 4 = 0

-x + y – 12 = 0

Slope of the polar is – \(\frac{a}{b}=\frac{1}{1}\) = 1

Distance from the centre

Question 7.

If ax + by + c = 0 is the polar of (1, 1) with respect to the circle x² + y² – 2x + 2y +1=0 and H.C.F. of a, b, c is equal to one then find a² + b² + c².

Solution:

Equation of the circle is

x2+ y2-2x + 2y +1 = 0

Polar of (1, 1) w.r.to the circle is

x.1 + y.1 -(x + 1) + (y + 1) + 1 = 0

x + y – x – 1 + y + 1 + 1 = 0

2y + 1 = 0 …………. (1)

Given equation of the line ax + by + c = 0 ……….. (2)

Comparing (1) and (2)

\(\frac{a}{0}=\frac{b}{2}=\frac{c}{1}\) = k

a = 0, b = 2k, c = k

a² + b² + c² = 0 + 4k² + k² = 5k²

H.C.F of (a, b, c) = 1 ⇒ k = 1

a² + b² + c2 = 5(1)² = 5

III.

Question 1.

Find the coordinates of the point of in-tersection of tangents at the points where x + 4y – 14 = 0 meets the circle x² + y² – 2x + 3y – 5 = 0.

Solution:

Equation of the given circle is

x² + y² – 2x + 3y – 5 = 0

polar of P(x_{1}, y_{1}) is

2xx_{1} + 2yy_{1} – 2x – 2x_{1} + 3y + 3y_{1} – 10 = 0

2(x_{1} – 1)x + (2y_{1} + 3)y – (2x_{1} – 3y_{1} + 10) = 0 ……….. (1)

Equation of QR is x + 4y – 14 = 0 ……… (2)

Comparing (1) and (2)

∴ Co-ordinates of p are (\(\frac{109}{76}\), \(\frac{9}{38}\))

Question 2.

If the polar of the points on the circle x² + y² = a² with respect to the circle x² + y² = b² touches the circle x² + y² = c² then prove that a,b,c are in Geometrical progression.

Solution:

P(x_{1}, y_{1}) is a point on the circle x² + y² = a²

∴ x²_{1} + y²_{1} = a² ……. (1)

Polar of P w.r.to the circle x²+ y² = b² is xx_{1} + yy_{1} = b²

There is a tangent to the circle x² + y² = c²

Cross – multiplying we get b2 = ac

∴ a, b, c are in Geometric progression.

Question 3.

Tangents are drawn to the circle x² + y² = 16 from the point P(3, 5). Find the area of the triangle formed by these tangents and the chord of contact of P.

Solution:

Equation of the circle is x² + y² = 16

Polar of P (3, 5) is 3x + 5y = 16

PL = Length of the perpendicular from P

\(\frac{|9+25-16|}{\sqrt{9+25}}=\frac{18}{\sqrt{34}}\)

Centre of the circle = C (0, 0)

P = Length of the perpendicular from c

Question 4.

Find the locus of the point whose polars with respect to the circles x² + y² – 4x – 4y – 8 = 0 and x² + y² – 2x + 6y – 2 = 0 are mutually perpendicular.

Solution:

Suppose P(x, y) is any position the locus Equation of the circles are

x² + y² – 4x – 4y – 8 = 0 ………….. (1)

x² + y² – 2x + 6y – 2 = 0 …………. (2)

Equation of the polar of P w.r.to circle (1) is

xx_{1} + yy_{1} – 2(x + x_{1}) – 2(y + y_{1}) – 8 = 0

x(x_{1} – 2) + y(y_{1} – 2) – (2x_{1} +2y_{1} + 8) = 0 ………… (3)

Polar of P w.r. to circle (2) is

xx_{1} + yy_{1} – 1(x + x_{1}) – 3(y + y_{1}) – 2 = 0

xx_{1} + yy_{1} – x – x_{1} + 3y + 3y_{1} – 2 = 0

x(x_{1} – 1) + y(y_{1} + 3) – (x_{1} + 3y_{1} + 2) = 0 ………. (4)

(3) and (4) are perpendicular 0

⇒ a_{1}a_{2} + b_{1}b_{2}

(x_{1} – 2) (x_{1} – 1) + (y_{1} – 2) (y_{1} + 3) = 0

x²_{1} – 3x_{1} + 2 + y²_{1} + y_{1} – 6 = Q

x²_{1} + y²_{1} – 3x_{1} + y_{1} – 4 = 0

Locus of p(x_{1}, y_{1}) is x² + y² – 3x+ y – 4 = 0

Question 5.

Find the locus of the foot of the perpen¬dicular drawn from the origin to any chord of the circle S ≡ x² + y² + 2gx + 2fy + c = 0 which subtends a right angle at the origin.

Solution:

Suppose L (x_{1}, y_{1}) is the foot of the per-pendicular from the origin on the chord QL

Slope a QL = \(\frac{y_{1}}{x_{1}}\)

Slope of QR = \(\frac{x_{1}}{y_{1}}\)

Equation of QR is y – y_{1} = \(\frac{x_{1}}{y_{1}}\)(x – x_{1})

yy_{1} – y²_{1} = -xx_{1} + x²_{1}

xx_{1} + yy_{1} = x²_{1} + y²_{1}

or \(\frac{xx_{1}+yy_{1}}{x_{1}^{2}+y_{1}^{2}}\) = 1 …………. (1)

Equation of the circle is

x² + y² + 2gx + 2fy + c = 0 …………. (2)

Hamogenising (2) with the help of (1), combined equation of

QQ, OR is x² + y² + (2gx + 2fy)

OQ, OR are perpendicular

Co – eff. of x² + co- eff of y² = 0

2(x²_{1} + y²_{1}) + 2gx_{1} + 2fy_{1} + c =

Locus of L (x_{1}, y_{1})is

2(x² + y²) + 2gx + 2fy + c = 0