If you’ve worked on a sufficient number of coordinate geometry problems, you can well understand how sometimes the solutions are incredibly long. And boring too! Most students think that all that there is to coordinate geometry is working with long and tedious expressions, and, in most cases, ending up with a wrong answer!

Well, some of this is true to an extent. Doing coordinate geometry problems does become a lengthy and tedious task at times. However, there are ways to simplify this task, the most important being the use of plane geometry wherever possible. We’re on a regular basis surprised by students who try to employ algebraic equations in a problem that can be solved in seconds using simple plane geometry facts.

In this article, we pick some problems from coordinate geometry at random, and show how an algebraic approach compares with a plane geometry approach in each case.

Problem – 1
There are two circles S1 and S2 and given by

From a point P on S2, tangents are drawn to S1 as shown. Find the angle between the tangents.

This problem is in fact a no-brainer: even a student very inexperienced in coordinate geometry would realize that plane geometry is the best way to solve this.

A coordinate approach would have involved assuming the coordinates of P, say

and writing the joint equation of the pair of tangents from P to S1. The angle can then be determined by using the standard appropriate relations on this joint equation.

A plane geometry approach is much simpler: join P to the center O of the circles and join O to one of the points of contact, say Q, as shown:Now, simply observe that


Thus,
or
so that
Thus, is two steps, we’re able to determine that the two tangents are perpendicular to each other.

Although this problem is very simple, it still manages to make a general point: pure geometry can be way better than an algebraic approach in many cases.

Problem – 2

There are two circles S1 and S2 given by

From a point P on S2 , a pair of tangents is drawn to S1. The triangle formed using these two tangents and the chord of contact QR, is completed. Where does the centroid of triangle PQR lie?

Note that the radius of S2 is twice that of S1.

Now, let us first write down what steps you’d have taken in case of an algebraic approach:

- assume coordinates for P

- write the equation to the pair of tangents, and thus determine the coordinates of Q and R

- write the coordinates for the centroid G using the coordinates of P, Q and R

- using these coordinates, figure out where G lies

Lengthy! Very lengthy indeed!

Instead, we follow a plane geometry approach, recalling from our class 9th studies that the centroid divides any median in the ratio 2:1.

First join P, Q and R to O, the center of the circles. Also, some more constructions are done, as shown:

Now starts the magic. First of all, note that angle(AOR)= 60⁰ and OR = OB = r, implying triangle ORB is equilateral, so that OA = AB.

Also, since OB = BP, we get BP = 2AB, implying that B divides PA in the ratio 2:1. Finally, note that PA is the median, implying that B is the centroid of triangle PQR.

Thus, the centroid of triangle PQR lies on the inner circle. How easy was this?

Problem – 3

There are two circles S1 and S2, with S1 lying completely internal to S2 . Find the locus of a point P which moves such that it is always equidistant from S1 and S2.

A coordinate geometry approach would involve the following:

- assume the equations for S1 and S2 (warning! – note that these two circles are not necessarily concentric)

- assume an arbitrary point P(x,y), and write its distances from S1 and S2. If we denote the radii of the two circles by r₁ and r₂, we have

distance(P,S1) = PC₁ – r₁

distance(P,S2) = r₂ - PC₂ ….(1)

- write the coordinate expressions for these two distances

- once done, equate the two distances and simplify to get the locus of P.

This is again very lengthy.

Using the plane geometry approach, we are done simply after we’ve written (1):

PC₁ – r₁ = r₂ - PC₂

–> PC₁ + PC₂ = r₁ + r₂

–> P moves in such a way that  PC₁ + PC₂ is a constant equal to  r₁ + r₂

–> P moves on an ellipse! And the foci are C1 and C2

You see that apart from determining the locus of P to be an ellipse, we have succeeded in showing that the foci of this ellipse are C1 and C2 . All in two lines!

As an exercise, redo this problem in case the two circles were external to each other.

There three examples should have given you a brief and basic idea on how plane geometry is used to solve coordinate geometry problems. If you intend to become a good math student, you must always use your intuition to realize the most appropriate approach to a problem: algebraic, plane geometry or a mixture of the two. This skill can be developed only with practice.

We’ll bring you more such problems in the future.