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Show that the common tangents to the cir...

Show that the common tangents to the circles `x^2+y^2-6x=0` and `x^2+y^2+2x=0` from an equilateral triangle.

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We have parabola `y^(2)=4x` and circle `(x+1)^(2)+y^(2)=1`.
As shown in the figure, circle touches the parabola at origin.
So, one of the common tangents is y-axis.
The other two common tangents are symmetrical about x-axis.
Let equation of tangent of tangent AB to parabola be
`y=mx+(1)/(m)`
Solving it with the circle, we have
`x^(2)+(mx+(1)/(m))^(2)+2x=0`
`or" "(1+m^(2))x^(2)+4x+(1)/(m^(2))=0`
Since line touches the circle, above equation has two equal roots.
`:." Discriminant", D=0`
`rArr" "16-(4)/(m^(2))(1+m^(2))=0`
`rArr" "4m^(2)=1+m^(2)`
`rArr" "m=pm(1)/(sqrt(3))`
`i.e.," "angleBAO=angleOCA=(pi)/(6)`
Hence the triangle is equilateral.
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