Question

The equation of the circle passing through the points (-2,4) and (6,4) and
whose centre lies on the line 3x-4y-2=0 is​

Answer 1

$\huge \bold \red{ \fcolorbox{green}{grey}{required \: answer}}$

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through points (-2,4) and (6,4)

(4 – h)2 + (1 – k)2 = r2 … (1)

(6 – h)2 + (4 – k)2 = r2 … (2)

Since the centre (h, k) of the circle lies on line 3x-4y -2 = 0

-2h + k = 4 … (3)

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

(-2 – h)2 + (4 – k)2 = (6 – h)2 + (4 – k)2

⇒ 4 + 4h + h2 + 16 – 8k + k2 = 36 – 12h + h2 + 16– 8k + k2

⇒4 – 4h + 16 – 8k = 36 – 12h + 16 – 8k

⇒ -4h - 12h - 8k + 8k = 36+16-16-4

⇒-16h = 52-20

⇒-16h = 32

$\implies \: h = \frac{32}{ - 16}$

⇒h = -2

now substitute the value of h in equation 3

-2h + k = 4

⇒-2(-2) + k = 4

⇒ 4 + k = 4

⇒k = 4 - 4

⇒k = 0

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

(4 – h)² + (1 – k)² = r² …(1)

(4 + 2)² + (1 – 0)² = r² … (1)

⇒ (6)² + (1)² = r²

⇒ 26 + 1 = r²

⇒ r² = 27

$\bold { \tt{⇒ \: r = \sqrt{27} }}$

Thus, the equation of the required circle is

( x - h )² + ( y - k )² = r²

(x – [-2] )² + (y –0)² =(√27)²

x² + 4x + 4 + y² - 0y + 0 = 27

x² + y² + 4x + 4 = 27

x² + y² + 4x = 23

x² + y² + 4x - 23 = 0

Answer 2

Answer:

$\huge \fbox \red{answer}$

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through points (-2,4) and (6,4)

(4 – h)2 + (1 – k)2 = r2 … (1)

(6 – h)2 + (4 – k)2 = r2 … (2)

Since the centre (h, k) of the circle lies on line 3x-4y -2 = 0

-2h + k = 4 … (3)

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

(-2 – h)2 + (4 – k)2 = (6 – h)2 + (4 – k)2

⇒ 4 + 4h + h2 + 16 – 8k + k2 = 36 – 12h + h2 + 16– 8k + k2

⇒4 – 4h + 16 – 8k = 36 – 12h + 16 – 8k

⇒ -4h - 12h - 8k + 8k = 36+16-16-4

⇒-16h = 52-20

⇒-16h = 32

\begin{gathered} \implies \: h = \frac{32}{ - 16} \\ \end{gathered}

⟹h= 32/-16

⇒h = -2

now substitute the value of h in equation 3

-2h + k = 4

⇒-2(-2) + k = 4

⇒ 4 + k = 4

⇒k = 4 - 4

⇒k = 0

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

(4 – h)² + (1 – k)² = r² …(1)

(4 + 2)² + (1 – 0)² = r² … (1)

⇒ (6)² + (1)² = r²

⇒ 26 + 1 = r²

⇒ r² = 27

Thus, the equation of the required circle is

( x - h )² + ( y - k )² = r²

(x – [-2] )² + (y –0)² =(√27)²

x² + 4x + 4 + y² - 0y + 0 = 27

x² + y² + 4x + 4 = 27

x² + y² + 4x = 23

x² + y² + 4x - 23 = 0