Number of Common Tangents to Two Circles in Different Conditions
The equations...
Question
The equations of circles with radius 3 units and touching the circle x2+y2−2x−4y−20=0 at (5,5) is/are
A
(5x−16)2+(5y−13)2=225
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B
(5x−13)2+(5y−16)2=225
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C
(5x−34)2+(5y−37)2=225
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D
(5x−37)2+(5y−34)2=225
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Solution
The correct options are B(5x−13)2+(5y−16)2=225 D(5x−37)2+(5y−34)2=225
Given circle is x2+y2−2x−4y−20=0 C=(1,2),r=5
As seen from figure, there are 2 possible circles of 3 units, Now, slope of the line joining centres tanθ=5−25−1=34⇒cosθ=45,sinθ=35
Using parametric form of line The coordinates of C1 and C2 are =(5±3cosθ,5±sinθ)=(5±3(45),5±3(35))⇒C1=(135,165),C2=(375,345)
∴ The required equations of the circles are (x−135)+(y−165)=9⇒(5x−13)2+(5y−16)2=225(x−375)+(y−345)=9⇒(5x−37)2+(5y−34)2=225
Alternate solution: Given circle is x2+y2−2x−4y−20=0 C=(1,2),r=5 Let the centre of the circle be (h,k) Now, k−5h−5=5−25−1⇒k−5h−5=34⋯(1) And (h−5)2+(k−5)2=32 Using equation (1), we get ⇒25(h−5)216=9⇒h−5=±125⇒h=135,375⇒k=165,345
So, the equation of circles are (x−135)+(y−165)=9⇒(5x−13)2+(5y−16)2=225(x−375)+(y−345)=9⇒(5x−37)2+(5y−34)2=225