If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, – 1), then the radius of C is :
JEE · Math · previous-year question
- A.5correct
- B.2$$\sqrt {5} $$
- C.4
- D.$$\sqrt {37} $$
Answer
A. 5
Explanation
x2 + y2 + 4x $$-$$ 6y $$-$$ 12 = 0 Equation of tangent at (1, $$-$$ 1) x $$-$$ y + 2(x + 1) $$-$$ 3(y $$-$$ 1) $$-$$ 12 = 0 3x $$-$$ 4y $$-$$ 7 = 0 $$ \therefore $$ Equation of circle is (x2 + y2 + 4x $$-$$ 6y $$-$$ 12) + $$\lambda $$ (3x $$-$$ 4y $$-$$ 7) = 0 It passes through (4, 0) : (16 + 16 $$-$$ 12) + $$\lambda $$ (12 $$-$$ 7) = 0 $$ \Rightarrow $$ 20 + $$\lambda $$(5) = 0 $$ \Rightarrow $$ $$\lambda $$ = $$-$$ 4 $$ \therefore $$ (x2 + y2 + 4x $$-$$ 6y $$-$$ 12) $$-$$ 4(3x $$-$$ 4y $$-$$ 7) = 0 or x2 + y2 $$-$$ 8x + 10y + 16 = 0 Radius = $$\sqrt {16 + 25 - 16} = 5$$
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