I.S.I. 2015 subjective solution (Problem 1)

Let \(y = x^2 + ax + b \) be a parabola that cuts the coordinate axes at three distint points. Show that the circle passing through these three points also passes through (0,1).


Teacher: There are two alternative routes of solving this problem. One is using the general equation of a circle. Second is by computing the center of the circle.

Student: Let me try them one by one. We take the general equation to a circle \(x^2 + y^2 + 2gx + 2fy + c = 0 \)

Suppose this circle passes through the three given points in which the parabola \(x^2 + ax + b\) cuts the coordinate axes. We put y = 0 in equation to circle to get, \(x^2 + 2gx + c \). Surely this equation cuts the x axis at two points at which the parabola cuts the x -axis. As the leading coefficient in both equations is 1, hence we may compare the coefficients to conclude 2g = a and c = b

Now we rewrite the equation to the circle replacing 2g by a and c by b to have \(x^2 + y^2 + ax + 2fy + b = 0 \)

In the equation \(y = x^2 +ax + b \) putting x = 0 we find the y intercept as (0, b). Hence circle also passes through (0, b)

Hence \((0)^2 + b^2 + a*0 + 2fb + b = 0 \) .

Since b is not 0 (then the parabola \(x^2 + ax = x(a+x) \) would have intersected the coordinate axes at two points only that is (0,0) and (-a, 0) ) we have 2f = – (b+1).

Hence the final equation to circle is \(x^2 + y^2 + ax -(b+1)y + b \). This equation is definitely satisfied by (0, 1). Therefore the circle passes through (0,1) as well.

Teacher: Excellent. Now let us quickly sketch the aliter.

The two points at which parabola intersects x axis are \(\displaystyle { (\frac{-a + \sqrt{a^2 – 4b} }{2} , 0 ) , (\frac{-a – \sqrt{a^2 – 4b} }{2} , 0) } \) . The perpendicular bisector of these two points is x = -a/2

Similarly the perpendicular bisector of (0, b) and (0, 1) is y = (b+1)/2

We take the intersection of these two lines (-a/2, (b+1)/2 ) and show that it is the center (by measuring distance from the vertices).

4 Replies to “I.S.I. 2015 subjective solution (Problem 1)”

  1. Aaaaa….we can easily convert it into a simple geometric problem….
    (i) suppose the roots are a_1 and a_2. then just take all the four points and apply ptolemy’s theorem for general quadrilateral, it is easy to show that the equality holds. Hence, cyclic.
    (ii) We can use simple coordinate geometry concept to compute the slopes of some lines joining the 4 points. Then by applying some basic cyclic quadrilateral properties, the result will be obvious! :

  2. Sir, what about if we rewrite the parabola as (x-E)*(x-F)? Three points will be then (0,EF), (E,0), (F,0). we take last two as end points of a diameter and get the equation of the circle. Satisfying with the third one we get some condition putting which we get the equation satisfied by (0,-1).

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