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Internal common tangents of x^2+y^2-4x-4y+4=0 and x^2+y^2+6x+6y+9=0 - JEE Advanced Circle

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

Question
Given the circles x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0 and x2+y2+6x+6y+9=0x^2+y^2+6x+6y+9=0, their internal common tangents are
Ax−y=2x-y=2 and x+2y=3x+2y=3
B2x−3y=42x-3y=4 and x+5=5x+5=5
Cx=0x=0 and y=0y=0correct
Dx+1=0x+1=0 and y−3=0y-3=0
Solution
Question attachment Step 1:
x2+y2−4x−4y+4=0:C1=(2,2),r1=4+4−4=2x^2+y^2-4x-4y+4=0:\quad C_1=(2,2),\quad r_1=\sqrt{4+4-4}=2
x2+y2+6x+6y+9=0:C2=(−3,−3),r2=9+9−9=3x^2+y^2+6x+6y+9=0:\quad C_2=(-3,-3),\quad r_2=\sqrt{9+9-9}=3
 Step 2:
C1C2=52+52=52≈7.07>r1+r2=5C_1C_2 = \sqrt{5^2+5^2} = 5\sqrt2 \approx 7.07 > r_1+r_2 = 5
 so each circle lies outside the other and exactly two internal common tangents exist. Step 3: C1=(2,2)C_1=(2,2) is at distance 22 from both axes and r1=2r_1=2; C2=(−3,−3)C_2=(-3,-3) is at distance 33 from both axes and r2=3r_2=3. So x=0x=0 and y=0y=0 are tangent to both circles. Step 4: The two circles lie on opposite sides of each axis, so each axis separates them, and a common tangent that separates two circles is an internal one. Being two in number, x=0x=0 and y=0y=0 are the two internal common tangents. Equivalently, the internal tangents meet on C1C2C_1C_2 at the point dividing it internally in the ratio r1:r2=2:3r_1:r_2 = 2:3, which is (3⋅2+2⋅(−3)5,3⋅2+2⋅(−3)5)=(0,0)\left(\frac{3\cdot 2+2\cdot(-3)}{5},\frac{3\cdot 2+2\cdot(-3)}{5}\right)=(0,0), and x=0, y=0x=0,\ y=0 are the two lines through the origin. Answer: (3) x=0x=0 and y=0y=0
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