What will be the equation of the circle which touches the line x + 2y + 5 = 0 and passes through the point of intersection of the circle x^2 + y^2 = 1 and x^2 + y^2 + 2x + 4y + 1 = 0?

(a) x^2 + y^2 + 2x + y = 0

(b) x^2 + y^2 + x + 2y = 1

(c) x^2 + y^2 + x + 2y = 0

(d) x^2 + y^2 + 2x + 2y = 1

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The above asked question is from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12

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The correct answer is (c) x^2 + y^2 + x + 2y = 0

Explanation: The equation of any circle through the points of intersection of the given circle is,

x^2 + y^2 + 2x + 4y + 1 + k(x^2 + y^2 – 1) = 0

x^2 + y^2 + 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0

Clearly, the co-ordinates of the center of the circle (1) are, (-1/(1 + k), -2/(1 + k)) and its radius,

= √[(1/(1 + k))^2 + (2/(1 + k))^2 – ((1 – k)]/(1 + k))

= √(4 + k^2)/(1 + k)

Clearly, the line x + 2y + 5 = 0 is tangent to the circle (1), hence, the perpendicular distance of the line from the center of the circle = radius of the circle

± (-1/(1 + k))– 2(2/(1 + k)) + 5/ √(1^2 + 2^2) = √(4 + k^2)/(1 + k)

Or ±(5k/√5) = √(4 + k^2)

Or 5k^2 = 4 + k^2

Or 4k^2 = 4

Or k = 1 [as, k ≠ -1]

Putting k = 1 in (1), equation of the given circle is,

x^2 + y^2 + x + 2y = 0