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Radius of circle touching (x-2)^2+y^2=1 and the bisectors of the transverse tangents - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Given that a circle touching (x−2)2+y2=1(x-2)^2+y^2=1 and also the angle bisectors of the transverse common tangents of (x±2)2+y2=1(x\pm2)^2+y^2=1 has radius p±qp\pm\sqrt q, the value of p+qp+q is
A88
B99correct
C1010
D1111
Solution
Step 1: The circles (x−2)2+y2=1(x-2)^2+y^2=1 and (x+2)2+y2=1(x+2)^2+y^2=1 have centres (±2,0)(\pm2,0) and radii 11, so they are symmetric about the origin and their internal common tangents pass through it. For y=mxy=mx,
∣2m−0∣m2+1=1  ⟹  4m2=m2+1  ⟹  m=±13\frac{|2m-0|}{\sqrt{m^2+1}} = 1 \;\Longrightarrow\; 4m^2 = m^2+1 \;\Longrightarrow\; m = \pm\frac{1}{\sqrt3}
 so the transverse common tangents are y=x3y=\dfrac{x}{\sqrt3} and y=−x3y=-\dfrac{x}{\sqrt3}. Step 2: These make +30∘+30^\circ and −30∘-30^\circ with the xx-axis, so they are symmetric about the xx-axis and about the yy-axis, and the bisectors are
y=0andx=0y = 0 \quad\text{and}\quad x = 0
 the two coordinate axes. Step 3: A circle of radius rr touching both axes has centre (±r,±r)(\pm r,\pm r), and its distance from (2,0)(2,0) does not depend on the sign of the ordinate. With abscissa −r-r, external and internal tangency to (x−2)2+y2=1(x-2)^2+y^2=1 give
(r+2)2+r2=(r+1)2  (r+2)^2+r^2 = (r+1)^2 \;
⟹  r2+2r+3=0,(r+2)2+r2=(r−1)2  ⟹  r2+6r+3=0\Longrightarrow\; r^2+2r+3 = 0, \qquad (r+2)^2+r^2 = (r-1)^2 \;\Longrightarrow\; r^2+6r+3 = 0
 with no positive root, and internal tangency with abscissa +r+r gives
(r−2)2+r2=(r−1)2  ⟹  r2−2r+3=0(r-2)^2+r^2 = (r-1)^2 \;\Longrightarrow\; r^2-2r+3 = 0
 with no real root. So the centre is (r,r)(r,r) or (r,−r)(r,-r) and the touch is external. Step 4:
(r−2)2+r2=r+1\sqrt{(r-2)^2+r^2} = r+1
(r−2)2+r2=(r+1)2(r-2)^2+r^2 = (r+1)^2
r2−4r+4+r2=r2+2r+1r^2-4r+4+r^2 = r^2+2r+1
r2−6r+3=0⟹r=6±36−122=3±6r^2-6r+3 = 0 \quad\Longrightarrow\quad r = \frac{6\pm\sqrt{36-12}}{2} = 3\pm\sqrt6
 Both roots are positive, so both are genuine radii. Step 5:
r=3±6=p±q  ⟹  p=3, q=6  ⟹  p+q=9r = 3\pm\sqrt6 = p\pm\sqrt q \;\Longrightarrow\; p=3,\ q=6 \;\Longrightarrow\; p+q = 9
 Answer: (2) 99
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