Question

Find the equation of circle whose centre is (3,-1) and which cutts off an intercept of length 6 from the line 2x-5y+18=0

Answer 1

Equation of the line cutting the circle : 2x - 5y + 18 = 0

The perpendicular drawn to the line from the centre of the circle will bisect the line since the radii of the circle and the line form a isosceles triangle.

We need to find the distance between the centre of the circle to the point where the perpendicular meets the line.

We need to find the equation of the perpendicular first :

Writing the equation of the line is Slope-Intercept form :

2x - 5y + 18 = 0

5y = 2x + 18

y = (2/5)x + 18/5

Slope = 2/5

So the slope of the line perpendicular to this line is -5/2.

The equation of the perpendicular line is :

y = mx + b

y = (-5/2)x + b

This line passes through the point (3, -1)

-1 = (-5/2)*3 + b

b = 15/2 - 1

b = 13/2

So the equation of the perpendicular line is :

y = (-5/2)x + 13/2

2y = -5x + 13

2y + 5x - 13 = 0

To find the point at which these 2 lines meet we need to solve the equations,

5x + 2y - 13 = 0

2x - 5y + 18 = 0

multiplying first equation by 5 and second by 2

25x + 10y - 65 = 0

4x - 10y + 36 = 0

Adding both equations,

29x - 29 = 0

x = 1

replacing value of x in first equation,

5 + 2y -13 = 0

2y - 8 = 0

2y = 8

y = 4

So the line meet at the point (1, 4)

The distance between this point and the centre is

√((1-3)² + (4+1)²) = √29

The perpendicular drawn to the line forms a right angles triangle with the line with radius as the hypotenuse.

Now we need calculate the radius of the circle,

r² = (√29)² + 3²

r² = 29 + 9

r² = 38

r = √38

No we have both centre of the circle and its radius, the equation of the circle is :

(x-3)² + (y+1)² = (√38)²

x² - 6x + 9 + y² + 2y + 1 = 38

x² + y² - 6x + 2y - 28 = 0