Conic Sections

16. Centre (0, 2) and radius 2 .

Here, h = 0, k = 2, r = 2

The equation of the circle is given by'

(x – h)² + (y – k)² = r²

x² + (y – 2)² = 4

$\Rightarrow x^2+y^2-4y=0$

13. Let the equation of the circle be,

(x – h)² + (y – k)² = r² – 0-----(i)

As the circle is passing through (0, 0) we get,

(0 – h)² + (0 – k)² = r²

(–h)² + (–k)² = r²

h² + k² = r²--------(ii)

The circle also makes intercepts a and b on the cooperate axes.

Let intercept on x–axis be 'a' and on y–axis be 'b

12. Given,

r = 5.

Since the centre lies on x – axis.

k = 0.

? Centre of circle = (h, k) = (h, 0)

The equation of the circle in given by,

(x – h)² + (y – k)² = r²

(x – h)² + (y – 0)² = (5)²

(x – h)² + y² = 25- (i)

Since the curie parses through the point (2, 3),

Putting x = 2 and y = 3 in equation (1),

11. Let the equation of the circle be.

(x – h)² + (y – k)² = r² -----(i)

Since the circle passes through (2, 3) and (–1, 1),

Putting x = 2 and y = 3 in (i),

(2 – h)² + (3 – k)² = r²---------(ii)

Putting x = –1 and y = 1 in (i),

(–1 – h)² + (1 – k)² = r²

Equating equation (ii) and (iii), We get:–

(2 – h)² +

10. Let the equation of the circle be,

(x – h)² + (y – k)² = r² - (i)

Since the circle passes through (4, 1) and (6, 5)

Putting x = 4 and y = 1 in (i),

(4 – h)² + (1 – k)² = 2² - (ii)

Putting x = 6 and y = 5 in (i),

(6 – h)² + (5 – k)² = r²- (iii)

Equating equation (ii) and (iii), We get.

(4 – h)²

9. 2x² + 2y²= x

2(x² + y²) = x

x² + y² = $\frac{x}{2}$

x² – $\frac{x}{2}$ + y²= 0

x² – 2 $\left(\frac{x}{4}\right)$ + y² = 0

x² – $2\left(\frac{x}{4}\right)+{\left(\frac{1}{4}\right)}^2-{\left(\frac{1}{4}\right)}^2+y^2=0$

x² – $2\left(\frac{1}{4}x\right)+{\left(\frac{1}{4}\right)}^2+y^2={\left(\frac{1}{4}\right)}^2$

Using,

(a – b)² = a² + b² – 2ab,

We get

${\left(x-\frac{1}{4}\right)}^2+{(y-0)}^2={\left(\frac{1}{4}\right)}^2$

Comparing with the equation of a circle (x – h)² + (y – k)² = r²

We get,

h = $\frac{1}{4}$ , k = 0, r = $\frac{1}{4}$

?Centre = (h, k) = $\left(\frac{1}{4},0\right)$

Radian = r = $\frac{1}{4}$

3. Given,

h = $\frac{1}{2}$ , k = $\frac{1}{4}$ , r = $\frac{1}{12}$

? Equation of the circle is,

(x – h)² + (y – k)² = r²

${\left(x-\frac{1}{2}\right)}^2+{\left\{y-\left(\frac{1}{4}\right)\right\}}^2={\left(\frac{1}{12}\right)}^2$

${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{1}{4}\right)}^2=\frac{1}{144}$

$\Rightarrow 144x{}_2+144y^2-144x-72y+44=0$

2. Centre (–2, 3) and radius 4.

Given, h = –2, k = 3, r = 4

? Equation of the circle is,

(x – h)² + (y – k)² = 2²

(x + 2)² + (y – 3)² = 16

$\Rightarrow x^2+y^2+4x-6y-3=0$