The circles ${x^2} + {y^2} - 10x + 16 = 0$ and ${x^2} + {y^2} = {r^2}$ intersect each other at two distinct points if:
A.r < 2
B. r > 8
C.2 < r < 8
D.2 < r < 9
VerifiedHint: We will first draw the figure of both the circles in XY – plane. We will calculate the radius of the first circle using the formula of radius:
r = ${(x – coordinate\, of\, centre)}^{2} + ({y – coordinate\, of\, centre)}^{2} – constant term^{\dfrac{1}{2}}.$
And, for coordinates of the centre, C$ \equiv ${(– coefficient of x/2), (– coefficient of y/2)}. The centre of the second circle will origin as there is no term of x. Then, we will check the cases where both of the circles will intersect at two distinct points on the graph. We will then obtain the range of the value of r i.e., the restriction over the region in which r lies.
Complete step-by-step answer:
We are given two circles: ${x^2} + {y^2} - 10x + 16 = 0$and ${x^2} + {y^2} = {r^2}$.
Let us see the first circle: ${x^2} + {y^2} - 10x + 16 = 0$
The centre of this circle can be calculated as: C$ \equiv ${(– coefficient of x/2), (– coefficient of y/2)}. Since, coefficient of x = - 10 and coefficient of y = 0, therefore
$ \Rightarrow $C $ \equiv $(5, 0)
Now, the radius of this circle will be given by the formula:
r = ${(x – coordinate\, of\, centre)}^{2} + ({y – coordinate\, of\, centre)}^{2} – constant term^{\dfrac{1}{2}}.$
$ \Rightarrow $
r = $\sqrt {{5^2} + 0 - 16} = \sqrt {25 - 16} = \sqrt 9 = 3$
Now, the radius of the first circle is 3 units.
Let us look at the second circle: ${x^2} + {y^2} = {r^2}$
Here, as we can see that the coefficients of x and y are 0, hence, the centre will be at origin (0, 0).
Its radius is r.
Now, we can plot these two circles in the XY – plane as shown below. We will consider all those situations where both the circles intersect each other at two distinct points.
In the above graph, the orange circle represents the first circle ${x^2} + {y^2} - 10x + 16 = 0$having radius 3 and the blue circles represents the second circle ${x^2} + {y^2} = {r^2}$with the conditions where both the circles intersect at two distinct points.
So, from the figure, we can deduce that the minimum radius of the second circle${x^2} + {y^2} = {r^2}$ will be r > 2 since when it intersects, it exceeds the radius 2 (see figure).
The maximum radius that the second circle${x^2} + {y^2} = {r^2}$ can have is r < 8 because when the bigger circle (blue) intersects, it doesn’t exceed the radius 8 (see figure).
Combining both of the conditions (inequalities), we can say that the radius of the second circle ${x^2} + {y^2} = {r^2}$lies between 2 < r < 8.
Therefore, option(C) is correct.
Note: In such questions, you may get confused while deducing the radius of the first circle from its centre coordinates because they are also required to be calculated.
You may get wrong while plotting the conditions for the value of the radius of the second circle ${x^2} + {y^2} = {r^2}$ because it is mentioned that both the circles intersect at two distinct points.
You can plot the bigger circle (blue) anywhere else but make sure that both the circles intersect exactly at two distinct points and the centre of the circle is at origin.
