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Question: If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10p...

If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10p, then the equation of the circle is-

A

x2 + y2 – 2x + 2y – 23 = 0

B

x2 + y2 – 2x – 2y – 23 = 0

C

x2 + y2 + 2x + 2y – 23 = 0

D

x2 + y2 + 2x – 2y – 23 = 0

Answer

x2 + y2 – 2x + 2y – 23 = 0

Explanation

Solution

Solving 2x + 3y + 1 = 0 and 3x – y –4 = 0

we get centre (1, –1)

As circumference = 10p

2pr = 10p ή r = 5

required equation of circle is

(x – 1)2 + (y + 1)2 = 52