Question

find the equation of circle which passes through two points on the x-axis which are at distance 4 from the origin and whose radius is 5 ​

Answer 1

Step-by-step explanation:

Given  

find the equation of circle which passes through two points on the x-axis which are at distance 4 from the origin and whose radius is 5

The general equation of circle is x^2 + y^2 + 2gx + 2fy + c = 0 ------1

The two points on x-axis at a distance of 4 units from origin can be (4,0) and (-4,0)

So equation 1 passes through these two points.

So (4,0) will be 4^2 + 0^2 + 2g.4 + 2f.0 + c = 0

                  16 + 8 g + c = 0------------2

Now (-4,0) will be (-4)^2 + 0^2 + 2g.(-4) + 2f.0 + c = 0

                              16 – 8 g + c = 0---------3

Adding 2 and 3 we get

                              32 + 2c = 0

So c = - 32 / 2

Or c = - 16

From 2 we have

16 + 8 g – 16 = 0

8 g = 0

Or g = 0

Now radius = 5

So √g^2 + f^2 – c = 5

√0^2 + f^2 – (-16) = 5

Squaring we get,

So f^2 + 16 = 25

So f^2 = 9

Or f = ± 3

Substituting in 1 we get

x^2 + y^2 + 2x.0 + 2y(±3) – 16 = 0

So x^2 + y^2 ±6y – 16 = 0 is the required equation.