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Incident ray reflected from the tangent at (0,-2) to x^2+y^2=4 - find J + K - JEE Advanced Circle

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

Question
A ray of light strikes the point (−3,−2)(-3,-2) on the tangent drawn at (0,−2)(0,-2) to the circle x2+y2=4x^2+y^2=4, and after reflection it touches that circle. Given that the incident ray moved along the line Jx−Ky+46=0Jx-Ky+46=0, find the value of J+KJ+K.
Solution
Answer: 7
Question attachment Step 1: The tangent to x2+y2=4x^2+y^2=4 at (0,−2)(0,-2) is 0⋅x+(−2)y=40\cdot x + (-2)y = 4, i.e. the horizontal line y=−2y=-2, which is the mirror. Step 2: The reflected ray is a tangent from P(−3,−2)P(-3,-2) to the circle. The vertical line x=−3x=-3 is at distance 33 from the centre, so it is not a tangent and the reflected ray has a slope. A line through PP is y+2=m(x+3)y+2 = m(x+3), i.e. mx−y+3m−2=0mx-y+3m-2=0, and tangency to the circle of centre (0,0)(0,0), radius 22 requires
∣3m−2∣1+m2=2\frac{|3m-2|}{\sqrt{1+m^2}} = 2
 Step 3: Squaring,
9m2−12m+4=4+4m2  ⟹  5m2−12m=0  ⟹  m(5m−12)=09m^2-12m+4 = 4+4m^2 \;\Longrightarrow\; 5m^2-12m = 0 \;\Longrightarrow\; m\left(5m-12\right)=0
m=0orm=125m = 0 \quad\text{or}\quad m = \frac{12}{5}
 m=0m=0 is the mirror y=−2y=-2 itself, so the reflected ray has slope 125\dfrac{12}{5}. Step 4: The mirror is horizontal, so reflection in it reverses the sign of the slope, and the incident ray through P(−3,−2)P(-3,-2) has slope −125-\dfrac{12}{5}.
y+2=−125(x+3)  ⟹  5y+10=−12x−36  ⟹  12x+5y+46=0y+2 = -\frac{12}{5}(x+3) \;\Longrightarrow\; 5y+10 = -12x-36 \;\Longrightarrow\; 12x+5y+46 = 0
 Step 5:
Jx−Ky+46=0≡12x+5y+46=0  ⟹  J=12,−K=5  ⟹  K=−5Jx-Ky+46 = 0 \equiv 12x+5y+46=0 \;\Longrightarrow\; J = 12,\quad -K = 5 \;\Longrightarrow\; K=-5
J+K=12+(−5)=7J+K = 12+(-5) = 7
 Answer: 77
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