Question

(a) Find the centre and radius of the circle whose equation is

(i) \( (x-2)^{2}+(y+3)^{2}=9 \)

(ii) \( 2 x^{2}+2 y^{2}-6 x+10 y=1 \) 

(iii) \( x^{2}+y^{2}+8 x-2 y+13=0 \) 

(b) (i) Find the Cartesian equation of the curve \( \frac{2}{r}=1+\cos \theta \) 

(ii) Find the polar equation of \( x^{2}+y^{2}=9 \) 

(c) (i) Find the equation of a straight line which passes through the two points \( (-2,3) \) and \( (2,-5) \) 

(ii) Find the equation of the tangent to the circle \( x^{2}+y^{2}+2 x+4 y-3=0 \) at the point \( (1,-4) \)

Answer 1

(a) (i) Given equation of circle is

(x - 2)² + (y + 3)² = 9

⇒ (x - 2)² + y(-(-3))² = 3²

\(\therefore\) Centre of circle is (2, -3) and radius of the circle is 3.(But comparing the standard form of circle whose centre is (h, k) and radius is r; i.e; (x - h)² + (y - k)² = r²)

(ii) Given equation of circle is

2x² + 2y² - 6x + 10y = 1

⇒ x² + y² - 3x + 5y = 1/2 (Dividing both sides by 2)

Compare with x² + y² + 2fx + 2gy + c = 0

we get 2f = -3 ⇒ f = -3/2

2g = 5 ⇒ g = 5/2

& c = -1/2

\(\therefore\) Centre of the circle = (-f, -g) = (3/2, -5/2)

and radius of the circle is r = \(\sqrt{f^2+g^2-c}\) 

⇒ r = \(\sqrt{(\frac{-3}2)^2+(5/2)^2-(-1/2)}\)

 = \(\sqrt{\frac94+\frac{25}4+\frac12}\) 

 = \(\sqrt\frac{9+25+2}4\) = \(\sqrt\frac{36}4=\sqrt9\) = 3

\(\therefore\) Radius of the circle is r = 3

& centre of the circle is(-3/2, 5/2)

(iii) Given equation of circle is 

x² + y² + 8x - 2y + 13 = 0

⇒ x² + 8x + 16 - 16 + y² - 2y + 1 - 1 + 13 = 0

⇒ (x + 4)² + (y - 1)² + 13 - 17 = 0

⇒ (x - (-4))² + (y - 1)² = 2²

\(\therefore\) Centre of the circle is(-4, 1)

and radius of the circle is r = 2

(b)(i) Polar equationo of the curve is

\(\frac2r=1+cos\theta\)----(1)

\(\because\) x = r cos θ and y = r sin θ

\(\therefore\) cos θ = x/r

Then equation(1) converts into

2/r = 1 + x/r

⇒ 2 = x + r

⇒ 2 - x = r

⇒ r² = (2 - x)²

⇒ x² + y² = x² - 4x + 4 (\(\because\) r² = x² + y²)

⇒ y² = -4x + 4 which is a cartesian equation of the given curve.

(ii) Given cartesian equation of the curve is

x² + y² = 9

⇒ r² = 9 (\(\because\) x² + y² = r²)

⇒ r = 3

(c) (i) Equation of line which passes through two points (-2, 3) & (2, -5) is

\(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

⇒ y - 3 = \(\frac{-5-3}{2-(-2)}(x - (-2))\) 

⇒ 2x + y + 1 = 0 which is equation of required line.

(ii) Given equation of curve is

x² + y² + 2x + 4y - 3 = 0----(1)

By diferentiating equation (1) w.r.t. x

we get

2x + 2y dy/dx + 2 + 4 dy/dx = 0

⇒ (2y + 4)dy/dx = -(2x + 2)

⇒ dy/dx = \(\frac{-(2x+2)}{2y+4} = \frac{-(x+1)}{y+2}\)

\(\therefore\) slope of tangent to curve (1) at point (1, -4) is

m = (dy/dx)(1, -4) = \(\frac{-(1+1)}{-(4+2)}=\frac{-2}{-2}=1\)

\(\therefore\) Equation of tangent to curve (1) at point (1, -4) is y-(-4) = m(x - 1)

⇒ y + 4 = 1(x - 1)

⇒ x - y = 5

which is equation of tangent to given curve at point(1, - 4)