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Exercise 2(b) · Q4

Q.Find the equation of the radical axis of the circles x2+y2+4x−7=0x^2+y^2+4x-7=0 and 2x2+2y2+3x+5y−9=02x^2+2y^2+3x+5y-9=0.

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Step 1. S1:x2+y2+4x−7=0S_1:x^2+y^2+4x-7=0 is already in standard form (g=2,f=0,c=−7g=2,f=0,c=-7).

Step 2. S2:2x2+2y2+3x+5y−9=0S_2:2x^2+2y^2+3x+5y-9=0 has leading coefficient 22, not 11, so it must be normalised (divided by 22) before it can be subtracted from S1S_1:

x2+y2+32x+52y−92=0.x^2+y^2+\frac32x+\frac52y-\frac92=0.

Step 3. Now subtract: S1−S2(norm)=(4x−7)−(32x+52y−92)S_1-S_2^{\text{(norm)}}=\Big(4x-7\Big)-\Big(\frac32x+\frac52y-\frac92\Big). …

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