Correct Answer - Option 3 : 5
From question, the equation of circle given is:
⇒ x2 + y2 + 4x – 6y = 12
The equation on the tangent to the circle is given by the equation:
⇒ xx1 + yy1 + 2(x + x1) – 3(y + y1) – 12 = 0
From question, x1 = 1 and y1 = -1,
⇒ x – y + 2(x + 1) – 3(y – 1) – 12 = 0
∴ 3x – 4y – 7 = 0
The equation of the circle which is touching the above tangent at point (1,-1) is:
⇒ (x – 1)2 + (y + 1)2 + λ(3x – 4y – 7) = 0 ----(1)
Where λ ∈ R
The required circle passes through the point (4, 0).
⇒ (4 – 1)2 + (0 + 1)2 + λ(3 × 4 – 4 × 0 – 7) = 0
⇒ 9 + 1 + λ(5) = 0 ⇒ λ = -2
On substituting λ = -2 in Equation (1),
⇒ (x – 1)2 + (y + 1)2 ± 2(3x – 4y – 7) = 0
∴ x2 + y2 – 8x + 10y + 16 = 0
The general equation of the circle is given as:
x2 + y2 + 2gx + 2fy + c = 0
On comparing general and obtained equation,
⇒ g = -4; f = 5; c = 16
The radius is given by the formula:
\({\rm{Radius}} = \sqrt {{g^2} + {f^2} - c}\)
On substituting the required values,
\(\Rightarrow {\rm{Radius}} = \sqrt {16 + 25 - 16}\)
∴ Radius = 5
