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Tangents at ends of the chord of x^2+y^2-6x-4y+3=0 with mid point (1,1) - JEE Advanced Circle

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

Question
Let ABAB be a chord of the circle x2+y2−6x−4y+3=0x^2+y^2-6x-4y+3=0 whose mid point is (1,1)(1,1). The tangents drawn at AA and BB meet at the point
A(1,0)(1,0)
B(−1,0)(-1,0)correct
C(0,1)(0,1)
D(1,−1)(1,-1)
Solution
Step 1: S=x2+y2−6x−4y+3S = x^2+y^2-6x-4y+3 and the mid point is (1,1)(1,1). The chord with a given mid point is T=S1T=S_1.
T=x+y−3(x+1)−2(y+1)+3=−2x−y−2T = x+y-3(x+1)-2(y+1)+3 = -2x-y-2
S1=1+1−6−4+3=−5S_1 = 1+1-6-4+3 = -5
 So ABAB is
−2x−y−2=−5⟹2x+y−3=0-2x-y-2 = -5 \quad\Longrightarrow\quad 2x+y-3 = 0
 Step 2: Let the tangents meet at (x1,y1)(x_1,y_1). Its chord of contact is T=0T=0.
xx1+yy1−3(x+x1)−2(y+y1)+3=0xx_1+yy_1-3(x+x_1)-2(y+y_1)+3 = 0
(x1−3) x+(y1−2) y+(−3x1−2y1+3)=0(x_1-3)\,x + (y_1-2)\,y + \left(-3x_1-2y_1+3\right) = 0
 Step 3: This is the same line as 2x+y−3=02x+y-3=0, so
x1−32=y1−21=−3x1−2y1+3−3\frac{x_1-3}{2} = \frac{y_1-2}{1} = \frac{-3x_1-2y_1+3}{-3}
 Step 4: From the first pair, x1−3=2(y1−2)x_1-3 = 2(y_1-2), i.e. x1=2y1−1x_1 = 2y_1-1. Substituting in the second pair,
−3(y1−2)=−3(2y1−1)−2y1+3=−8y1+6-3(y_1-2) = -3(2y_1-1)-2y_1+3 = -8y_1+6
−3y1+6=−8y1+6⟹5y1=0⟹y1=0-3y_1+6 = -8y_1+6 \quad\Longrightarrow\quad 5y_1 = 0 \quad\Longrightarrow\quad y_1 = 0
 and x1=−1x_1 = -1. Step 5: Power of (−1,0)(-1,0) with respect to the circle:
1+0+6−0+3=10>01+0+6-0+3 = 10 > 0
 so the point lies outside and two real tangents do meet there. Answer: (2) (−1,0)(-1,0)
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