Let’s start with an example:

Let R represent x^{2} + (y-3)^{2} – 4 = 0. Does R = Z have a real solution?

- Z is represented by y = 3
- Z is represented by mx + y = 3

Readers are encouraged to take some time to try to solve the above – challenge yourself to 30 seconds 🙂

**Trick: For questions involving determination of real solutions, it is sometimes quicker to plot rough sketches of the equations on the xy axis**

**There are two critical ‘two variables’ equations we should get super familiar with**

**The first one ax + by = c represents a straight line on the xy axis.**

**The second: (x – a) ^{2} + (y – b)^{2} = d^{2} represents a circle on the xy axis with center (a,b) and radius of d**

R seems pretty similar to our circle equation

R can be rewritten as x^{2} + (y-3)^{2} = 2^{2} (a circle with center at 0,3 and radius of 2 units)

Let’s plot this…

Now (1) tells us that Z is represented by y = 3.

y = 3 passes through the center our of circle and any line passing through the center of its circle will intersect its perimeter at some points – yielding real solutions as shown below.

Bingo! So (1) by itself is sufficient

Option (2) seems a bit more complex. We have an additional variables m and x to consider. Now, regardless of the value of m, the value of mx = 0 at x = 0.

So, at x = 0, mx + y = 3 simplifies to y = 3 so mx + y = 3 always passes through 0,3 which is the center of our circle.

Hence (2) by itself is also sufficient.

Hence, we go with **option (D)** 🙂