A circle always passes through the fixed points (a, 0) and (–a, 0).Q.The radius of the circle x2 + y2 + x + c = 0 passing through the origin isa)1/4b)1/2c)1d)2Correct answer is option 'B'. Can you explain this answer?

Question

A circle always passes through the fixed points (a, 0) and (&#8211;a, 0).Q.The radius of the circle x2 + y2 + x + c = 0 passing through the origin isa)1/4b)1/2c)1d)2Correct answer is option 'B'. Can you explain this answer?

Answer

Given equation of circle is x^2 + y^2 + x c = 0

To find the radius of the circle passing through origin, we need to find the distance between the origin and the center of the circle.

Let (h, k) be the center of the circle.

Since the circle passes through (a, 0), we can use the distance formula to get:

√((h-a)^2 + k^2) = |h-a|

Since the circle also passes through (-a, 0), we can use the distance formula again to get:

√((h+a)^2 + k^2) = |h+a|

Squaring both equations, we get:

(h-a)^2 + k^2 = (h+a)^2 + k^2

Simplifying, we get:

h = a/2

Substituting this value of h in either equation, we get:

k = √(3a^2/4)

The distance between the origin and the center of the circle is:

r = √(h^2 + k^2) = √(a^2/4 + 3a^2/4) = √(a^2) / 2 = a/2

Therefore, the radius of the circle passing through the origin is a/2.

The correct answer is option B) 1/2.

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