Question

find the equation of the circle which passes through (4,1),(6,5) and having centre on 4x+y-24=0

Answer 1

Answer:

The equation of circle is

$(x-\frac{37}{7})^2+(y-\frac{20}{7})^2=\frac{250}{49}$

Step-by-step explanation:

Let (a,b) be the center and r  be the radius of unknown circle then as per given conditions, the distance of each of points  (4,1), (6,5) from center will be equal to radius  r  as

$(a-4)^2+(b-1)^2=r^2$ →  (1)

$(a-6)^2+(b-5)^2=r^2$ → (2)

(1) and (2) becomes

$a^2+b^2-8a-2b+17=r^2$

$a^2+b^2-12a-10b+61=r^2$

Subtracting above two, we get

4a+8b=44 ⇒ a+2b=11 → (3)

Since, the center  (a,b) of circle lies on the straight line  4x+y-24=0, hence it will satisfy the equation of straight line i.e

4a+b=24 →(4)

Solving (3) and (4), we get

$b=\frac{20}{7}$ and $a=\frac{37}{7}$

which gives $r^2=\frac{250}{49}$

Hence, the the equation of circle becomes

$(x-a)^2+(y-b)^2=r^2$

⇒ $(x-\frac{37}{7})^2+(y-\frac{20}{7})^2=\frac{250}{49}$

Answer 2

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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.