Question

find the equation of the circle whose centre lies on the line 4x+y=16 and which passes through the points (4,1) and (6,5).​

Answer 1

Step-by-step explanation:

Refer to attachment..........

Answer 2

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Answer:

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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.