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A circle of radius r such that both coordinates of its centre are positive, touches the x-axis and line 3y - 4x = 0, then the equation of circle is
 

Option: 1

x ^{2} + y ^{2} + 4rx - 2ry + 4r^{2} = 0


Option: 2

x ^{2} + y ^{2} - 4rx - ry + 4r^{2} = 0


Option: 3

x ^{2} + y ^{2} -2rx -2 ry + r^{2} = 0

 


Option: 4

x ^{2} + y ^{2} -4rx -2 ry +4 r^{2} = 0


Answers (1)

Answer (4)

Circle touches x-axis so its y-co-ordinate of x-axis = radius = r Now it touches line 3y - 4x = 0

So,

\frac{3r-4h}{5}=\pm r\\ \Rightarrow 3r-4h=5r\: or \: 3r - 4h = -5r\\ \Rightarrow h= \frac{-2r}{4}\; \; \; \; \; h=2r

h should be positive

So, h = 2r

Now equation circle is

\left ( x-2r \right )^{2}+\left ( y-r \right )^{2}=r^{2}

x ^{2} + y ^{2} -4rx -2 ry +4 r^{2} = 0

Posted by

Sumit Saini

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