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Exercise 2(a) · Q1

Q.Find the angle between the circles x2+y2−4x−6y−3=0x^2+y^2-4x-6y-3=0 and x2+y2+2x+2y−2=0x^2+y^2+2x+2y-2=0.

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Step 1. For S:x2+y2−4x−6y−3=0S:x^2+y^2-4x-6y-3=0: g=−2,f=−3,c=−3g=-2,f=-3,c=-3, so centre C1=(2,3)C_1=(2,3) and r12=g2+f2−c=4+9+3=16r_1^2=g^2+f^2-c=4+9+3=16, i.e. r1=4r_1=4.

Step 2. For S′:x2+y2+2x+2y−2=0S':x^2+y^2+2x+2y-2=0: g′=1,f′=1,c′=−2g'=1,f'=1,c'=-2, so centre C2=(−1,−1)C_2=(-1,-1) and r22=1+1+2=4r_2^2=1+1+2=4, i.e. r2=2r_2=2.

Step 3. Distance between centres: d2=(2−(−1))2+(3−(−1))2=32+42=9+16=25d^2=(2-(-1))^2+(3-(-1))^2=3^2+4^2=9+16=25, so d=5d=5.

Step 4. Apply the angle formula:

cos⁡θ=d2−r12−r222r1r2=25−16−42(4)(2)=516.\cos\theta=\frac{d^2-r_1^2-r_2^2}{2r_1r_2}=\frac{25-16-4}{2(4)(2)}=\frac{5}{16}.

Step 5 (check via the coefficient form). cos⁡θ=c+c′−2gg′−2ff′2r1r2=−3−2−2(−2)(1)−2(−3)(1)16=−5+4+616=516\cos\theta=\dfrac{c+c'-2gg'-2ff'}{2r_1r_2}=\dfrac{-3-2-2(-2)(1)-2(-3)(1)}{16}=\dfrac{-5+4+6}{16}=\dfrac5{16}. Matches.

✓Final answer

The angle between the circles is θ=cos⁡−1(516)\theta=\cos^{-1}\Big(\dfrac5{16}\Big).

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