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Let P,Q,R and S be the feet of the perpe...

Let `P,Q,R` and `S` be the feet of the perpendiculars drawn from point `(1,1)` upon the lines `y=3x+4,y=-3x+6` and their angle bisectors respectively. Then equation of the circle whose extremities of a diameter are `R` and `S` is

A

`3x^(2)+3y^(2)+104x-110=0`

B

`x^(2)+y^(2)+104x-110=0`

C

`x^(2)+y^(2)-18x-4y+16=0`

D

`3x^(2)+3y^(2)-4x-18y+16=0`

Text Solution

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The correct Answer is:
D

`/_RBS=/_RAS=90^(@)`
so, equation of circle with `RS` as diameter is same circle with `AB` as diameter
So, point of intersection of two given lines
`y=3x+4`
and `y=-3x+6` is `y=6,x=1/3`
Equation of circle is
`(x-1)(x-1/3)+(y-1)(y-5)=0`
`implies3x^(2)+3y^(2)-4x-18y+16=0`
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