MathematicsStandard and General Equation of a CircleJEE Advanced 1978Moderate
Visualized Solution (Hindi)
Analyze the Given Circle
- Given Circle: x2+y2−2x−4y−20=0
- Comparing with x2+y2+2gx+2fy+c=0:
- Center C1(−g,−f)=(1,2)
- Radius r1=12+22−(−20)=25=5
The Geometric Condition
- Required Circle: Radius r2=5
- Point of Contact: P(5,5)
- Since r1=r2=5, P is the midpoint of the line joining C1(1,2) and C2(α,β).
Finding the New Center C2
- Using Midpoint Formula:
- 21+α=5⟹1+α=10⟹α=9
- 22+β=5⟹2+β=10⟹β=8
- New Center C2=(9,8)
Formulating the Final Equation
- Equation of Circle: (x−9)2+(y−8)2=52
- Expanding: (x2−18x+81)+(y2−16y+64)=25
- Simplifying: x2+y2−18x−16y+145−25=0
- Final Answer: x2+y2−18x−16y+120=0
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