Question

area,
(Ans. (x - 3)
25. Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre lies on the line
x - 3y - 11 = 0.
[Ans. x2 + y2 - 7x - 5y + 14 = 0]
x-3y - 1120​

Answer 1

EXPLANATION.

Equation of the circle passing through the point (2,3) and (-1,1).

Center lies on the line ⇒ x - 3y - 11 = 0.

As we know that,

General equation of the circle,

⇒ x² + y² + 2gx + 2fy + c = 0.

⇒ (x - h)² + (y - k)² = r².

Circle passing through points (2,3).

⇒ Let x = 2  and  y = 3,

Put the value of x and y in equation, we get.

⇒ (2 - h)² + (3 - k)² = r².

⇒ (4 + h² - 4h) + (9 + k² - 6k) = r².

⇒ 4 + h² - 4h + 9 + k² - 6k = r².

⇒ h² + k² - 4h - 6k + 13 = r² ⇒ (1).

Circle passing through points (-1,1).

⇒ Let x = -1  and  y = 1.

Put the value of x and y in equation, we get.

⇒ (-1 - h)² + (1 - k)² = r².

⇒ (1 + h² + 2h) + (1 + k² - 2k) = r².

⇒ 1 + h² + 2h + 1 + k² - 2k = r².

⇒ h² + k² + 2h - 2k + 2 = r². ⇒ (2).

Centre lies on the equation of line x - 3y - 11 = 0.

Let (h, k) be the points on the line x - 3y - 11 = 0.

⇒ h - 3k - 11 = 0. ⇒ (3).

From equation (1), (2) and (3) we get.

Solve equation (1) and (2) we get.

⇒ h² + k² - 4h - 6k + 13 = r² ⇒ (1).

⇒ h² + k² + 2h - 2k + 2 = r² ⇒ (2).

Equate the equation, we get.

⇒ h² + k² - 4h - 6k + 13 = h² + k² + 2h - 2k + 2.

⇒ - 4h - 6k + 13 = 2h - 2k + 2.

⇒ - 4h - 2h - 6k + 2k + 13 - 2 = 0.

⇒ - 6h - 4k + 11 = 0.

⇒ 6h + 4k - 11 = 0 ⇒ (4).

Solving equation (3) and (4) we get.

⇒ h - 3k - 11 = 0 ⇒ (3).

⇒ 6h + 4k - 11 = 0 ⇒ (4).

From equation (3) we get.

⇒ h = 3k + 11 ⇒ (5).

Put the value of equation (5) in equation (4) we get.

⇒ 6[3k + 11] + 4k - 11 = 0.

⇒ 18k + 66 + 4k - 11 = 0.

⇒ 22k + 55 = 0.

⇒ k = - 55/22.

⇒ k = - 5/2.

Put the value of k in equation (5), we get.

⇒ h = 3(-5/2) + 11.

⇒ h = -15/2 + 11.

⇒ h = - 15 + 22/2.

⇒ h = 7/2.

Put the value of (h, k) in equation (1), we get.

⇒ h² + k² - 4h - 6k + 13 = r².

⇒ (7/2)² + (-5/2)² - 4(7/2) - 6(-5/2) + 13 = r².

⇒ 49/4 + 25/4 - 14 + 15 + 13 = r².

⇒ 74/4 + 14 = r².

⇒ 74 + 56/4 = r².

⇒ 130/4 = r².

⇒ r² = 65/2.

Put all the values in equation of circle, we get.

⇒ (x - h)² + (y - k)² = r².

⇒ (x - 7/2)² + (y -(-5/2))² = 65/2.

⇒ (x - 7/2)² + (y + 5/2)² = 65/2.

⇒ (x² + 49/4 - 7x) + (y² + 25/4 + 5y) = 65/2.

⇒ x² + y² - 7x + 5y + 49/4 + 25/4 = 65/2.

⇒ x² + y² - 7x + 5y + 74/4 = 65/2.

⇒ x² + y² - 7x + 5y = 65/2 - 74/4.

⇒ x² + y² - 7x + 5y = 130 - 74/4.

⇒ x² + y² - 7x + 5y = 56/4.

⇒ x² + y² - 7x + 5y = 14.

⇒ x² + y² - 7x + 5y - 14 = 0.  

Answer 2

Question -

✰ Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre lies on the line x - 3y - 11 = 0.

Given that -

✠ The circle passing through the points (2, 3) and (-1, 1) and whose centre lies on the line x - 3y - 11 = 0

To find -

✠ The equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre lies on the line x - 3y - 11 = 0

Solution -

✠ The equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre lie on the line x - 3y - 11 = 0 is ${\sf{\bold{\red{x^{2} \: + \: y^{2} \: - 7x \: - 5y \: + 14 \: = 0}}}}$

Full solution -

$\rule{150}{1}$

~ Firstly let us assume ${\tt{\bold{\green{x^{2} \: + y^{2} \: + 2rm \: + 2ps \: + h \: = 0}}}}$ as the equation of the given circle.

${\bold{\bf{Where,}}}$

✵ -r and -p is the centre of the circle.

✵ r² and p² -h is the radius here.

$\rule{150}{1}$

~ Now as we already see in the question that the circle pasing through the points (2, 3) and (-1, 1) and whose centre lies on the line..!

⚕️ As the circle pas from the point (2,3) henceforth, it's known to be Eq(1)

➙ 4 + 9 + 4r + 6p + h = 0

➙ 13 + 4r + 6p + h = 0

➙ 4r + 6p + h = 0 - 13

➙ 4r + 6p + h = -13 Eq(2)

⚕️ As the circle pas from the point (-1,1) henceforth, it's known to be Eq(1)

➙ -2r + 2p + h = -2 Eq(3)

$\rule{150}{1}$

~ Now let's subtract Eq(3) from Eq(2)

• Continuing

➙ 6r + 4p = -11 Eq(4)

$\rule{150}{1}$

~ x-3y-11 = 0 is the given center line of the given circle,

⚕️ Since, -r and -p make equation as centre

➙ -r + 3p = 11 Eq(5)

$\rule{150}{1}$

~ Now let's solve Eq(4) and Eq(5)

➙ r = -7/2

➙ p = 5/2

$\rule{150}{1}$

~ Now let's solve Eq(2)

➙ h = -14

~ Henceforth, x² + y² - 7x - 5y + 14 = 0 is the eq of the given circle.

$\rule{150}{1}$