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Is x + y = 0 the common chord or radical axis of two given circles - JEE Advanced Circle

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

Question
Let the two circles be S1≡x2+y2−4x−6y−12=0S_1 \equiv x^2+y^2-4x-6y-12=0 and S2≡x2+y2+6x+4y−12=0S_2 \equiv x^2+y^2+6x+4y-12=0, and let the line be L≡x+y=0L \equiv x+y=0. Then
ALL touches both S1S_1 and S2S_2
BLL is the common chord of S1S_1 and S2S_2correct
CLL is the radical axis of S1S_1 and S2S_2correct
DLL is perpendicular to the line of centres of S1S_1 and S2S_2correct
Solution
Step 1:
S1: C1=(2,3),r1=4+9+12=5S_1:\ C_1=(2,3),\quad r_1=\sqrt{4+9+12}=5
S2: C2=(−3,−2),r2=9+4+12=5S_2:\ C_2=(-3,-2),\quad r_2=\sqrt{9+4+12}=5
 Step 2: The radical axis is S1−S2=0S_1-S_2=0.
S1−S2=(−4x−6y)−(6x+4y)=−10x−10y=0⟹x+y=0S_1-S_2 = (-4x-6y)-(6x+4y) = -10x-10y = 0 \quad\Longrightarrow\quad x+y = 0
 which is exactly LL, so (3) is correct. Step 3:
C1C2=52+52=52≈7.07C_1C_2 = \sqrt{5^2+5^2} = 5\sqrt2 \approx 7.07
 and ∣r1−r2∣=0<52<r1+r2=10|r_1-r_2| = 0 < 5\sqrt2 < r_1+r_2 = 10, so the circles intersect in two points. The radical axis of two intersecting circles is their common chord, so (2) is correct. Step 4: The distance from C1C_1 to LL is ∣2+3∣2=52<5=r1\dfrac{|2+3|}{\sqrt2} = \dfrac{5}{\sqrt2} < 5 = r_1, so LL cuts S1S_1 in two points and cannot touch it. Hence (1) is false. Step 5: Slope of C1C2C_1C_2 is 3−(−2)2−(−3)=1\dfrac{3-(-2)}{2-(-3)} = 1, slope of LL is −1-1, and their product is −1-1. So L⊥C1C2L \perp C_1C_2 and (4) is correct. Answer: (2), (3) and (4)
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