Find the equation of a circle of radius 5 which is touching another circle x2+y2−2x−4y−20=0 at (5, 5).

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a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

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Given  circle  is=(3,−1)x2+y2−2x−4y−20=0             2g=−2       ⇒g=−1            2f=−4       ⇒f=−2∴  Centre  C1=(1,2)and  radius  r=g2+f2−c                             =1+4+20=5Let  the  centre  of  the  required  circle  be  (h,k).Clearly,  P  is  the  mid  of  C1C2∴        5=1+h2    ⇒h=9and  5=2+k2    ⇒k=8Radius  of  the  required  circle=5∴  Equation  of  the  circle  is                               (x−9)2+(y−8)2=(5)2⇒    x2+81−18x+y2+64−16y=25⇒x2+y2−18x−16y+145−25=0⇒           x2+y2−18x−16y+120=0Hence,  the  required  equation  is   x2+y2−18x−16y+120=0.

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