Question

find the equation of the circle passing through the point (2,3) and (-1,1) and whose center is on the line x-3y-11=0​

Answer 1

Answer:

Let the equation of circle be

x

2

+y

2

+2gx+2fy+c=0 .....(1)

where (−g,−f) is center and g

2

+f

2

−c is the radius.

Since, the circle passes through (2,3), so it satisfies eqn (1)

⇒4+9+4g+6f+c=0

⇒4g+6f+c=−13 .....(2)

Since, the circle passes through (−1,1), so it satisfies eqn (1)

⇒1+1−2g+2f+c=0

⇒−2g+2f+c=−2 .....(3)

Subtracting eqn (3) from eqn (2), we get

6g+4f=−11 ....(4)

Given center lies on the line x−3y−11=0

Since, center (−g,−f) satisfies this equation.

⇒−g+3f=11 ....(5)

Solving eqn (4) and (5), we get

g=

2

−7

,f=

2

5

Put this value in (2), we get

c=−14

Substituting these values in (1),

x

2

+y

2

−7x+5y−14=0

which is the equation of required circle.

Step-by-step explanation:

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Answer 2

Given:-

• The equation of the circles passing through the point (2, 3) and (-1, 1).
• The centre is on the line x - 3y - 11 = 0

To find:-

• Find the equation of the circle..?

Solutions:-

• Let the equation of the equation of the required circle be (x - h)² + (y - k)² = r²

Since the circle passes through points (2, 3) and (-1, 1).

=> (2 - h)² + (3 - k)² = r².........(i).

=> (-1 - h)² + (1 - k)² = r².........(ii).

Since the centre (h, k) of the circle lies on line => x - 3y - 11 = 0

=> x - 3y = 11 ..........(iii).

From equation (i). and (ii), we obtain

=> (2 - h)² + (3 - k)² = (-1 - h)² + (1 - k)²

=> 4 - 4h + h² + 9 - 6k + k² = 1 + 2k + h² + 1 - 2k + k²

=> 4 - 4h + 9 - 6k = 1 + 2k + 1 - 2k

=> 6h + 4k = 11 ........(iv).

On solving equations (iii) and (iv), we obtain

• h = 7/2 and k = -5/2

On substitution the values of h and k in equation (i), we obtain.

=> (2 - 7/2)² + (3 - 5/2)² = r²

=> (4 - 7/2)² + (6 + 5/2)² = r²

=> (-3/2)² + (11/2)² = r²

=> 9/4 + 121/4 = r²

=> 130/4 = r²

So, the equation of the circle.

=> (x - 7/2)² + (y + 5/2)² = (130/4)²

=> (2x - 7/2)² - (2y + 5/2)² = 130/4

=> 4x² - 28x + 49 - 4y² + 20y - 56 = 0

=> 4x² + 4y² - 28x + 20y - 56 = 0

=> 4(x² + y² - 7x + 5y - 14) = 0

=> x² + y² - 7x - 5y - 14 = 0

Hence, the equation of the required circle is x² + y² - 7x - 5y - 14 = 0.