Select Page

This is I.S.I. 2015 Subjective 1 Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

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

## Discussion:

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).