Question

Find the equation of the circle passing through the Points (4, 1)and (6, 5) and whose centre is on the line is 4x + y = 16​

Answer 1

Given points,

(4,1),(6,5)

equation of circle (x−h)² + (y−k)² = r²

⇒(4−h)² + (1−k)² = r² ..(1)

⇒(6−h)² + (5−k)² = r² ....(2)

solving the above 2 equations, we get,

h + 2k = 11 ..(3)

given, 4h + k = 16 ...(4)

solving the above 2 equations, we get,

h = 3 , k = 4

substituting the above values in (1), we get,

(4−3)² + (1−4)² = r²

∴ r = 10

Hence, the equation is,

(x−3)² + (y−4)² = (√10)²

x² + y² − 6x − 8y + 15 = 0

Answer 2

$\huge\sf \pmb{\orange {\underline \pink{\underline{\:Ꭺ ꪀ \mathfrak ꕶ᭙ꫀя \: }}}}$

Let say the equation be (x – h)² + (y – k)² = r²

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

(4 – h)² + (1 – k)² = r² _______[1]

(6 – h)² + (5 – k)² = r²_______[2]

Since the centre (h, k) of the circle lies on line 4x + y = 16

4h + k = 16_______[3]

From equations (1) and (2), we obtain

(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²

16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²

16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

4h + 8k = 44

h + 2k = 11 _______[4]

On solving eqn (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in eqn (1), we obtain

(4 – 3)² + (1 – 4)² = r²

(1)² + (– 3)² = r²

1 + 9 = r²

r² = 10

r = √10

Thus, the equation of the required circle is

(x – 3)² + (y – 4)² = (√10)²

x² – 6x + 9 + y2² – 8y + 16 = 10

• x² + y² – 6x – 8y + 15 = 0

__________________________

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