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Circles drawn on the diameter as focal distance of any point lying on the parabola `x^(2)-4x+6y+10 =0` will touch a fixed line whose equation is a. y=1 b. y=-1 c. y=2 d. y=-2

Text Solution

Verified by Experts

The correct Answer is:
y+1=0

`x^(2)-4x+6y+10=0`
`or" "x^(2)-4x+4=-6-6y`
`or" "(x-2)^(2)=-6(y+1)`
The circle drawn on focal distance as diameter always touches the tangent drawn to the parabola at vertex.
Thus, the circle will touch the line y+1=0.
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