IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

kane

1.prove that the centre of a circle passing through (0,1) and touching the curve y=x² at (2,4) is (-16/5,53/10)

2.two circles are drawn through (a,5a) and (4a,a) touching the axis of y.prove that they intersect at an angle tan^-1 40/9

3.find the equation of circle which passes through (0,0) and cuts off chords of length b from the lines y=x and y= -x

ans:x²+y²+ root2 by=0 or x²+y²-root2 by=0 [by is not in the root]

4.show that the equation of image of circle x²+y²+16x-24y+183=0 by the line mirror 4x+7y+13=0 is x²+y²+32x+4y+235=0

5.a circle passes through (2,1) and the line x+2y=1 is a tangent to it at (3,-1).determine its equation.

ans:3(x²+y²)-23x-4y+35=0

plzzz explain the steps properly,5 points for one correct answer.

kane

plz yaar try atleast one of these

kane

somebody try yaar atleast one of them

chimanshu_007

HIII...

well the figure is at the end of the qs...

figure mai u can see....4 circles passing thru the origin (i have to adjust it a lil bit..) and they cut off = chords of length a frm straight line y = +_ x

o is at the centre where all the circles meet

angle AOB = BOC = COD = DOA = pi/2

so AB , BC , CD , DA are the diametres of 4 circles..

angle XOA = pi/4 and OA = b...

C₁A = bsin pi/4

OC₁ = bcos pi/4

so we get co-ordinates of A = [b/root2 , b/root2]

similarly co-ordinates of

B = [ -b/root2 , b/root2]

C = [-b/root2 , -b/root2]

D = [b/root2 , -b/root2]

now equation of circle wid AD as diameter...

(x - (b/root2))(x - (b/root2)) + (y - (b/root2))(y + (b/root2)) = 0

= x² + y² - (root2)bx = 0

similarly...u can find other 3 equations...

and there must b +_ in ur answer...

chimanshu_007

HERE'S THE FIGURE...MAY IT COME OUT CLEAR..

kane

thnxs a lot bro plz try the other questions too.

you have been rated according to my promise.

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

kane

thnxs a lot nadeemoidu

anyone for second

rooney

I have solved the 2nd Q earlier. Will try again.

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