please solve this question:

find the equation of the circle which passes through the points (1,-2) , (4,-3) and has its centre on the line 3x + 4y =7.

Question

please solve this question:
find the equation of the circle which passes through the points
(1,-2) , (4,-3) and has its centre on the line 3x + 4y =7

Lalit Mehra · Accepted Answer

Dear Student!

Let the centre of the circle be (h, k) and radius of the circle be r.

∴ Equation of the circle is (x – h)²+ (y – k)² = r² ...(1)

Given, (h, k) lies on the line 3x + 4y = 7

∴ 3h + 4k = 7 ...(2)

Given, (1, – 2) lies on (1).

∴ (1 – h)² + (– 2 – k)² = r²

⇒ (1 – h)²+ (2 + k)² = r² ...(3)

Given, (4, – 3) lies on (1).

∴ (4 – h)² + (– 3 – k)² = r²

⇒ (4 – h)² + (3 + k)² = r²...(4)

From (3) and (4), we have

${(1 - h)}^{2} + {(2 + k)}^{2} = {(4 - h)}^{2} + {(3 + k)}^{2} ∴ 1 + h^{2} - 2 h + 4 + k^{2} + 4 k = 16 + h^{2} - 8 h + 9 + k^{2} + 6 k \Rightarrow 6 h - 2 k = 20 \Rightarrow 3 h - k = 10 ... (5)$

Solving (2) and (5), we get

$3 h + 4 k = 7 3 h - k = 10 \underset{̲}{- + -} 5 k = - 3 ∴ k = - \frac{3}{5}$

When $k = - \frac{3}{5}$ , we have

$3 h - (\frac{- 3}{5}) = 10 (using (5)) ∴ 3 h = 10 - \frac{3}{5} = \frac{50 - 3}{5} \Rightarrow 3 h = \frac{47}{5} \Rightarrow h = \frac{47}{15}$

Using (4), we have

${(4 - \frac{47}{15})}^{2} + {(3 - \frac{3}{5})}^{2} = r^{2} ∴ r^{2} = {(\frac{60 - 47}{15})}^{2} + {(\frac{15 - 3}{5})}^{2} = (\frac{13}{15})^{2} + (\frac{12}{5})^{2} = \frac{169 + 144 \times 9}{225} = \frac{1465}{225}$

Equation of the circle is

${(x - \frac{47}{15})}^{2} + {[y - (- \frac{3}{5})]}^{2} = \frac{1465}{225} \Rightarrow (x - \frac{47}{15})^{2} + (y + \frac{3}{5})^{2} = \frac{1465}{15}$

Cheers!

90

please solve this question: find the equation of the circle which passes through the points (1,-2) , (4,-3) and has its centre on the line 3x + 4y =7.

please solve this question:

find the equation of the circle which passes through the points (1,-2) , (4,-3) and has its centre on the line 3x + 4y =7.