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Exercise 2(b) · Q5

Q.Show that the point (3,1)(3,1) lies on the radical axis of the circles x2+y2−4x−6y+11=0x^2+y^2-4x-6y+11=0 and x2+y2−2x−2y+1=0x^2+y^2-2x-2y+1=0, and show that the lengths of the tangents from this point to the two circles are equal.

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Step 1. For S:x2+y2−4x−6y+11=0S:x^2+y^2-4x-6y+11=0 and S′:x2+y2−2x−2y+1=0S':x^2+y^2-2x-2y+1=0, the radical axis is

S−S′=(−4x−6y+11)−(−2x−2y+1)=−2x−4y+10=0 ⟹ x+2y−5=0.S-S'=(-4x-6y+11)-(-2x-2y+1)=-2x-4y+10=0\ \Longrightarrow\ x+2y-5=0.

Step 2. Check that (3,1)(3,1) lies on this line: 3+2(1)−5=03+2(1)-5=0. Yes.

Step 3. Compute the power of (3,1)(3,1) with respect to SS: S(3,1)=9+1−12−6+11=3S(3,1)=9+1-12-6+11=3.

Step 4. Compute the power of (3,1)(3,1) with respect to S′S': S′(3,1)=9+1−6−2+1=3S'(3,1)=9+1-6-2+1=3. …

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