Question

Equation of the circle having diameter 8 and equation of diameters are  y + 2x - 5 = 0 and 2y + 3x - 8 = 0
1. x² + y² - 2x - 4y - 25 = 0
2. x2 + y2 - 4x - 2y - 11= 0
3. x2 + y2 + 4x + 2y - 25 = 0
4. x2 + y2 - 2x - 4y + 11 = 0

Answer 1

Correct Answer - Option 2 : x2 + y2 - 4x - 2y - 11= 0

Concept:

Standard equation of a circle:

\(\rm (x-h)^2+(y-k)^2=R^2\)

Where centre is (h, k) and radius is R.

Note: The intersection of the diameters is the centre of the circle.

 

Calculation:

Given diameter = 8

⇒ Radius = 4

Also diameters equations:

y + 2x - 5 = 0                    ...(i)

2y + 3x - 8 = 0                  ...(ii)

Substracting 2 × (i) from (ii)

-x + 2 = 0

x = 2

Putting back in (i)

y + 2 × 2 - 5 = 0

y = 1

So center is (2, 1) and radius 4

The equation of circle:

\(\rm (x-2)^2+(y-1)^2=4^2\)

⇒ \(\rm x^2 + 4 - 4x+y^2 + 1 - 2y = 16\)

⇒ \(\boldsymbol{\rm x^2 +y^2- 4x - 2y-11 = 0}\)