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 P1 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 S1 = 0
i.e., xx1 + yy1 + g(x + x1) + f(y + y1) + 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(x1, y1) w.r. to S = 0
xx1 + yy1 = 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 xx1 + yy1 = 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 S1 = 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 (x1, y1) be pole, the polar equation be
xx1 + yy1 – 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
xx1 + yy1 – 3(x + x1) – 4(y + y1) + 5 = 0
x(x1 – 3) + y(y1 – 4) – 3x1 – 4y1 + 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
xx1 + yy1 – $$\frac{5}{2}$$(x + x1) + 4(y + y1) + 6 = 0
(or) x(x1 – $$\frac{5}{2}$$) + y(y1 + 4) – $$\frac{5}{2}$$
x1 + 4y1 + 6 = 0 …………. (i)
x – 2y + 22 = 0 ………….. (ii)
(i) and (ii) both are equations of the same line comparing it we get

y1 = -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 (x1, y1) polar equation w.r.t.
x² + y² – 2x – 4y – 4 = 0 be
xx1 + yy1 -1 (x + x1) – 2(y + y1) – 4 = 0
x(x1 – 1) + y(y1 – 2) – x1 – 2y1 – 4 = 0 …….. (i)
2x + y + 5 = 0 is polar so comparing with equation (i) we get

x1 = -1, y1 = 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 l1x + m1y + n1 = 0; l2x + m2y + n2 = 0 are conjugate, S = 0 then,
r²(l1l2 + m1m2)
= (l1g + m1f – n1) (l2g + m2f – n2)
l1 = 1, m1 = 1, n1 = -5
l2 = 2, m2 = k, n2 = – 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
S1 ≡ x² + y² – 2x + 4y – 11 = 0
r = $$\sqrt{1+4+11}$$ = 4
S11 =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
S11 = 0 + 0 – 0 + 0 + 25 = 25 ⇒ √S11 = 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
Suppose θ (x1, y1) 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(x1, y1) be any point on the locus
S11 = x²1 + y²1 – a²

Squaring and cross-multiplying
a² = x² + y² – a²
x² + y² = 2a²
Locus of P (x1, y1) 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(x1, y1) is

2xx1 + 2yy1 – 2x – 2x1 + 3y + 3y1 – 10 = 0
2(x1 – 1)x + (2y1 + 3)y – (2x1 – 3y1 + 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(x1, y1) is a point on the circle x² + y² = a²
∴ x²1 + y²1 = a² ……. (1)
Polar of P w.r.to the circle x²+ y² = b² is xx1 + yy1 = 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
xx1 + yy1 – 2(x + x1) – 2(y + y1) – 8 = 0
x(x1 – 2) + y(y1 – 2) – (2x1 +2y1 + 8) = 0 ………… (3)
Polar of P w.r. to circle (2) is
xx1 + yy1 – 1(x + x1) – 3(y + y1) – 2 = 0
xx1 + yy1 – x – x1 + 3y + 3y1 – 2 = 0
x(x1 – 1) + y(y1 + 3) – (x1 + 3y1 + 2) = 0 ………. (4)
(3) and (4) are perpendicular 0
⇒ a1a2 + b1b2
(x1 – 2) (x1 – 1) + (y1 – 2) (y1 + 3) = 0
1 – 3x1 + 2 + y²1 + y1 – 6 = Q
1 + y²1 – 3x1 + y1 – 4 = 0
Locus of p(x1, y1) 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 (x1, y1) 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 – y1 = $$\frac{x_{1}}{y_{1}}$$(x – x1)
yy1 – y²1 = -xx1 + x²1
xx1 + yy1 = x²1 + y²1
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(x²1 + y²1) + 2gx1 + 2fy1 + c =
Locus of L (x1, y1)is
2(x² + y²) + 2gx + 2fy + c = 0