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
Answers
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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Answer:
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