Question

Find the equation to the circle which touches the axis of $$x$$ at a distance $$3$$ from the origin and intercepts a distance $$6$$ on the axis of $$y$$.

A
x2+y23x32y+5=0
B
x2+y26x32y+9=0
C
x2+y23x+32y+4=0
D
x2+y24x52y10=0

Solution

The correct option is B $$x^2+y^2-6x-3\sqrt{2}y+9=0$$The equation of the circle with centre $$(h,k)$$ and radius $$a$$ is$$(x-h)^2+(y-k)^2=a^2$$When the circle touches the x-axis the ordinate of the centre is equal to the radius of the circle i.e. $$k=a$$.Therefore, the equation becomes$$(x-h)^2+(y-a)^2=a^2$$$$x^2+y^2-2hx-2ay+h^2=0$$The circle passes through$$(3,0)$$ and the intercept made by a circle with $$y$$-axis is $$2\sqrt{f^2-c}$$So, $$2\sqrt{4a^2-h^2}=6$$     ......(1)and $$9-6h+h^2=0$$    ......(2)Solving (1) and (2), we get$$h=3$$$$a=\dfrac{3}{\sqrt{2}}$$So, the equation becomes$$x^2+y^2-6x-3\sqrt{2}y+9=0$$Maths

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