Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.

As the equation of the chord is,

2x – 5y + 18 = 0 ------- (i)

5y = 2x + 18

As, y = mx + C

Whereas, m is the slope of the line,

Slope of the line perpendicular to the chord,

As the product of slope of perpendicular lines = -1,

y - y₁ = m^' (x - x₁)

2y + 2 = -5x + 15

5x + 2y = 13 ------- (ii) [Equation of line passing from centre and cutting the chord]

Solving both the equations,

2x – 5y = -18 & 5x + 2y = 13

Multiplying the eq. (i) & eq. (ii) by 2 & 5 respectively,

4x – 10y = -36

25x + 10y = 65

29x = 29

x = 1

2(1) – 5y = -18

2 – 5y + 18 = 0

5y = 20

y = 4

Point of intersection @ chord and radius = (1,4)

Distance between the point of intersection & centre = [Distance Formula]

=√29 units

Using Pythagoras Theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

= (3)² + (√29)² = 29 + 9

= √38

Hypotenuse = √38 units (radius)

Since, the radius bisects the chord into two equal halves,

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – 3)² + (y –(-1))² = (√38)²

x² - 6x + 9 + (y + 1)² = 38

x² - 6x + y² + 2y + 1 + 9- 38 = 0

x² - 6x + y² + 2y – 28 = 0
Hence, the required equation of the circle is x² - 6x + y² + 2y – 28 = 0.