Question

: Find the equation of the circle passing through (4,1),(6,5) and having the centre on the line
4x + y-16=0.​

Answer 1

Answer:

hope ot helps you. have a nice day

Answer 2

$\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}$

Let the equation of the equation of the required circle 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______(i).

=> (6 - h)² + (5 - k)² = r²______(ii).

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

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

=> (4 - h)² + (1 - k)² = (6 - h)² + (5 - k)²

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

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

=> 4h + 8k = 44

=> h + 2k = 11 ______(iv).

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

On solving equations (iii) and (iv), we obtainh = 4 and k = 4.

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

=> (4 - 3)² + (1 - 4)² = r²

=> (1)² + (-3)² = r²

=> 1 + 9 = r²

=> 10 = r²

=> r = √10

So, the equation of the circle.

=> (x - 3)² + (y - 4)² = (√10)²

=> x² - 6x + 9 + y - 8y + 16 = 10

=> x² + y² - 6x - 8y + 15 = 0

Hence, the equation of the required circle is

x² + y² - 6x - 8y + 15 = 0.