The centre of the circle touching the circles x2+y2−4x−6y−12=0, x2+y2+6x+18y+26=0 at their point of contact and passing through the point (1,−1) is
We have two circles given by: ‌S1:x2+y2−4x−6y−12=0 ‌S2:x2+y2+6x+18y+26=0 We need to find a third circle that: Touches both S1 and S2 at their point of contact Passes through the point (1,−1) Any circle that touches both S1 and S2 at their point of contact can be written as: (x2+y2−4x−6y−12)+λ(x2+y2+6x+18y+26)=0 Since the circle also passes through the point (1,−1), substitute x=1,y=−1 into the equation: (1)2+(−1)2−4(1)−6(−1)−12+λ[(1)2+(−1)2+6(1)+18(−1)+26]=0 Calculate each term step-by-step: First bracket: 1+1−4−(−6)−12=2−4+6−12=−8 (Notice −6×−1=+6 ) Second bracket: 1+1+6−18+26=2+6−18+26=8−18+26=−10+26=16 So the equation becomes: −8+λ(16)=0 Solve for λ : λ=‌
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16
=‌
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Now plug λ=‌
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into our general circle equation: ‌(x2+y2−4x−6y−12)+‌
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(x2+y2+6x+18y+26)=0 ‌x2+y2−4x−6y−12+‌
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2
x2+‌
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y2+3x+9y+13=0 ‌(1+‌
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)x2+(1+‌
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)y2+(−4+3)x+(−6+9)y+(−12+13)=0
‌
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x2+‌
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y2−x+3y+1=0 To make it simpler, multiply both sides by 2 : 3x2+3y2−2x+6y+2=0 Divide all terms by 3 to get standard form: x2+y2−‌
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x+2y+‌
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=0 The center of a circle x2+y2+2gx+2fy+c=0 is (−g,−f). From the equation: ‌2g=−‌
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‌ so ‌g=−‌
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‌2f=2‌ so ‌f=1 Therefore, the center is (‌