Question

For how many values of p, the circle ${{x}^{2}}+{{y}^{2}}+2x+4y-p=0$and the coordinate axes have exactly three common points.

Answer 1

To solve this problem we need to divide it into three cases first when the circle passes through the origin and intersect x axis and y axis at one point each. Second case when the circle touches x axis and intersect y axis at two points and third case in which circle touches y axis and intersect x axis at two points. We will then calculate the values of p for each case using required conditions and count the values of valid p.

Complete step-by-step answer:
We need to find the number of values of p for which the circle ${{x}^{2}}+{{y}^{2}}+2x+4y-p=0$ and the coordinate axis have exactly three common points.
To solve this problem we will assume three cases, as
Case 1: when the given circle passes through origin and cuts the x and y axis at one point each.
We know that the circle passes through origin so, for this case it’s equation should satisfy the coordinate (0, 0).

So putting (x, y) = (0, 0) we get,
${{x}^{2}}+{{y}^{2}}+2x+4y-p=0$
$0+0+0+0-p=0$
$p=0$
Now we will see second case,
Case 2: When the given circle touches the x axis at a point other than origin and cuts the y axis at two distinct points.

We know that if the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ touches the x-axis and cuts the y-axis at two different point, then it should satisfy the following conditions,

& {{g}^{2}}-c=0 \\\ & {{f}^{2}}-c>0 \\\ \end{aligned}$$ We have equation as ${{x}^{2}}+{{y}^{2}}+2x+4y-p=0$, Now applying the above conditions for this case we get, g = 1, f = 2 and c = -p, $$\begin{aligned} & {{g}^{2}}-c=0 \\\ & \Rightarrow 1-(-p)=0 \\\ & \Rightarrow p=-1 \\\ & \\\ & {{f}^{2}}-c>0 \\\ & \Rightarrow 4-(-p)>0 \\\ & \Rightarrow p>-4 \\\ \end{aligned}$$ p = -1 is also greater than -4 so this is a valid value of p. Now third case, Case 3: When the given circle touches the y axis at a point other than origin and cuts the x axis at two distinct points.  Now for this case general conditions are same as case 2 but g and f interchanges in the conditions, So for this case we get, $$\begin{aligned} & {{f}^{2}}-c=0 \\\ & {{g}^{2}}-c>0 \\\ \end{aligned}$$ We have equation of circle as ${{x}^{2}}+{{y}^{2}}+2x+4y-p=0$, Applying the above conditions we get, g = 1, f = 2 and c = -p, $$\begin{aligned} & {{f}^{2}}-c=0 \\\ & \Rightarrow 4-\left( -p \right)=0 \\\ & \Rightarrow p=-4 \\\ & \\\ & {{g}^{2}}-c>0 \\\ & \Rightarrow 1-\left( -p \right)>0 \\\ & \Rightarrow p>-1 \\\ \end{aligned}$$ p = -4 is not greater than -1 so this is not a valid value of p. hence we get a total two values of p from case 1 and case 2 and that are 0 and -1. So there are 2 values of p for which circle ${{x}^{2}}+{{y}^{2}}+2x+4y-p=0$ and coordinate axes have exactly 3 common points. **Note:** You may make mistakes while solving case 2 and case 3 and apply only one condition to find the values of p in those cases and in that case you will end up getting a total 3 values of p, which is wrong. So you need to do that carefully and reject the invalid values of p according to the conditions.