IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

nadeemoidu

1)
the circle touches the curve y= x2 at (2,4) . i.e. the tangent of the curve and the circle at that point coincide. but radius is perpendicular to the tangent of the circle.

therefore, the normal to y=x2 at (2,4) passes thru the centre of the circle.

equation of this normal is 4y+ x=18 .............(1)

Now , the circle passes thru (0,1) and (2,4) .

So the centre passes thru the perpendicular bisector of these points. equation of this perpendicular bisector is 6y +4x =19 .........(2)

So the centre is the point of intersection of (1) and (2)

4)
centre of the circle is ( -8 ,12 ) and radius =5.

find the mirror image of ( -8 , 12 ) on 4x + 7y + 13 =0 . It is (-16,-2) .

now the reqd circle is ( x + 16 )^2 + (y+2 ) ^2 = 5^2

5)
x+2y =1 is a tangent to the circle at (3,-1) .
so the normal to the line at this point passes thru the centre of the circle.
equation of the normal is 2x-y=7 ....(1)

circle passes thru (2,1) and (3,-1) . so the centre passes thru the perpendicular bisector of these points. equation of this perpendicular bisector is 2x-4=5 ...(2)

So the centre of the circle is the point of intersection of (1) and (2) = (23/6 , 2/3)

radius = distance from (23/6 ,2/3) to (3,-1)

we get the reqd. circle

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