There are two circles whose equations are $$x2+y2=9and$$ $$x2+y2$$−$$8x$$−$$6y+n2=0$$, n∈Z having exactly two common tangents. The number of possible values of n, is

Analysis of two circles in circles

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There are two circles whose equations are $x^{2} + y^{2} = 9$
and $x^{2} + y^{2} - 8 x - 6 y + n^{2} = 0, n \in Z$ having exactly two common
tangents. The number of possible values of $n$ , is

Moderate

Solution

The coordinates of the centres and radii of the
circles are:
Centres: $C_{1} (0, 0) C_{2} (4, 3)$
Radii $r_{1} = 3 r_{2} = \sqrt{25 - n^{2}}, - 5 < n < 5$
Given circles will have exactly two common tangents, if
$\begin{array}{l} |r_{1} - r_{2}| < C_{1} C_{2} < r_{1} + r_{2} \\ \Rightarrow |3 - \sqrt{25 - n^{2}}| < 5 < 3 + \sqrt{25 - n^{2}} \end{array}$
Clearly, $|3 - \sqrt{25 - n^{2}}| < 5$ is true for all $n \in (- 5, 5)$
Now,
$\begin{aligned} 5 < 3 + \sqrt{25 - n^{2}} \\ \Rightarrow 2 < \sqrt{25 - n^{2}} \end{aligned}$
$\begin{array}{l} \Rightarrow 4 < 25 - n^{2} \\ \Rightarrow n^{2} - 21 < 0 \\ \Rightarrow - \sqrt{21} < n < \sqrt{21} \Rightarrow n = \pm 4, \pm 3, \pm 2, \pm 1, 0 . \end{array}$
Hence, $n$ can take $9$ integral values.

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Analysis of two circles in circles

Remember concepts with our Masterclasses.

There are two circles whose equations are x2+y2=9and x2+y2−8x−6y+n2=0, n∈Z having exactly two common tangents. The number of possible values of n, is

Ready to Test Your Skills?

free study material

There are two circles whose equations are $x^{2} + y^{2} = 9$
and $x^{2} + y^{2} - 8 x - 6 y + n^{2} = 0, n \in Z$ having exactly two common
tangents. The number of possible values of $n$ , is