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Tangents from (6,4) to x^2+y^2-4x-4y+4=0 meeting the y-axis: length AB - JEE Advanced Circle

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

Question
From the point (6,4)(6,4) two tangents are drawn to the circle x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0, and they cut the yy-axis at AA and BB. The length ABAB equals
A66
B88correct
C1010
D1212
Solution
Question attachment Step 1:
x2+y2−4x−4y+4=0  ⟹  (x−2)2+(y−2)2=22x^2+y^2-4x-4y+4=0 \;\Longrightarrow\; (x-2)^2+(y-2)^2 = 2^2
 centre C(2,2)C(2,2), radius 22. Step 2: The vertical line x=6x=6 is at distance 44 from CC, so it is not a tangent and every tangent from P(6,4)P(6,4) has a slope.
y−4=m(x−6)⟹mx−y+(4−6m)=0y-4 = m(x-6) \quad\Longrightarrow\quad mx-y+(4-6m)=0
 Step 3: Distance from the centre == radius:
∣2m−2+4−6m∣m2+1=2⟹∣2−4m∣=2m2+1\frac{|2m-2+4-6m|}{\sqrt{m^2+1}} = 2 \quad\Longrightarrow\quad |2-4m| = 2\sqrt{m^2+1}
 Step 4: Squaring,
4−16m+16m2=4m2+4  ⟹  12m2−16m=0  ⟹  4m(3m−4)=04-16m+16m^2 = 4m^2+4 \;\Longrightarrow\; 12m^2-16m = 0 \;\Longrightarrow\; 4m(3m-4)=0
m=0orm=43m = 0 \quad\text{or}\quad m = \frac{4}{3}
 Step 5: Putting x=0x=0. For m=0m=0 the tangent is y=4y=4, meeting the yy-axis at (0,4)(0,4). For m=43m=\tfrac43, y−4=43(0−6)y-4=\tfrac43(0-6) gives y=4−8=−4y = 4-8 = -4, i.e. (0,−4)(0,-4). Step 6:
AB=∣4−(−4)∣=8AB = |4-(-4)| = 8
 Answer: (2) 88
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