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The equation of circle with centre (4,3) and touching x^2+y^2=1 is

The equation of circle with centre (4,3) and touching x^2+y^2=1 is

Grade:12th pass

1 Answers

venkateshwarrao
17 Points
4 years ago
C1=(0,0)&r1=1,C2=(4,3) since the two circles touch each other externally then c1c2=r1+r2Implies 5=1+r2 implies r2=4Therefore the equation of the circle is(X-4)^2+(y-3)^2=16x^2+y^2-8x-6y+9=0

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