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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th Conic Sections

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

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Payal Gupta | Contributor-Level 10

4 months ago

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), We get.

(2 – h)² + (3)² = 25

2² + h² – 22.h + 9 = 25

4 + h²– 4h + 9 = 25

h² – 4h + 13 – 25 = 0

h² – 4h – 12 = 0

h² – (6 – 2)h – 12 = 0

h² – 6h + 2h – 12 – 0

h (h – 6) + 2 (h – 6) = 0

(h – 6) (h + 2) = 0

h = 6 and h = &

...more

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), We get.

(2 – h)² + (3)² = 25

2² + h² – 22.h + 9 = 25

4 + h²– 4h + 9 = 25

h² – 4h + 13 – 25 = 0

h² – 4h – 12 = 0

h² – (6 – 2)h – 12 = 0

h² – 6h + 2h – 12 – 0

h (h – 6) + 2 (h – 6) = 0

(h – 6) (h + 2) = 0

h = 6 and h = –2

When h = 6,

From (i), Equation of circle is,

(x –6)² + y² = 25

(x – 6)² + (y – 0)² = 52

When h = –2,

Equation of circle is, (x + 2)² + (y – 0)² = 5² or x²+2 $y^{2} + 4 x - 21 = 0$

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Slope of axis = $\frac{1}{2}$

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

⇒ 2y – 6 = x – 2

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2x + y – 6 = 0

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12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

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