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  • A foot of the normal from the point to a circle is <img alt="(2,1)" src="https://entrancecorner.oncodecogs.com/

A foot of the normal from the point (4,3) to a circle is (2,1)and a diameter of the circle has the equation 2 x-y=2. Then the equation of the circle is

Option: 1

x^2+y^2+2 x-1=0


Option: 2

x^2+y^2-2 x-1=0


Option: 3

x^2+y^2-2 y-1=0


Option: 4

None of these.


Answers (1)

best_answer

The line joining (4,3) and (2,1) is also along a diameter. So, the centre is the intersection of the diameters 2 x-y=2 and y-3=(x-4).Solving these, the centre =(1,0)

 \therefore \text { Radius }=\text { Distance between }(1,0) \text { and }(2,1)=\sqrt{2} \text {. }

\therefore$ Equation of circle $(x-1)^2+y^2=(\sqrt{2})^2 \Rightarrow x^2+y^2-2 x-1=0

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Anam Khan

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