The number of common tangents, to the circles x^2+y^2-18 x-15 y+131=0 and x^2+y^2-6 x-6 y-…

Question

The number of common tangents, to the circles $x^{2}+y^{2}-18 x-15 y+131=0$ and $x^{2}+y^{2}-6 x-6 y-7=0$, is :

Accepted Answer

We are given two circles:

(1) $x^2+y^2-18 x-15 y+131=0$

(2) $x^2+y^2-6 x-6 y-7=0$

First, let's find the centers and radii of the circles.

For circle (1):

Completing the square for the equation:

$(x^2-18x+{81})+(y^2-15y+\frac{225}{4})=-131+{81}+\frac{225}{4}$

$(x-9)^2+(y-\frac{15}{2})^2=\frac{25}{4}$

Center 1: $C_1(9, \frac{15}{2})$

Radius 1: $r_1 = \sqrt{\frac{25}{4}}=\frac{5}{2}$

For circle (2):

Completing the square for the equation:

$(x^2-6x+9)+(y^2-6y+9)=7+9+9$

$(x-3)^2+(y-3)^2=25$

Center 2: $C_2(3, 3)$

Radius 2: $r_2 = 5$

Now, let's find the distance between the centers :

$d = \sqrt{(9-3)^2 + (\frac{15}{2}-3)^2} = \sqrt{6^2 + \frac{9}{2}^2} = \sqrt{36 + \frac{81}{4}} = \frac{15}{2}$

Next, let's analyze the relative positions of the circles using the distance between centers and the sum and difference of the radii :

If $d > r_1 + r_2$, the circles are separate, and there are 4 common tangents.
If $d = r_1 + r_2$, the circles are externally tangent, and there are 3 common tangents.
If $0 < d < |r_1 - r_2|$, one circle is inside the other, and there are no common tangents.
If $d = |r_1 - r_2|$, the circles are internally tangent, and there is 1 common tangent.
If $d < |r_1 - r_2|$, one circle is completely inside the other, and there are no common tangents.

In this case :

$d = \frac{15}{2}$

$r_1 = \frac{5}{2}$

$r_2 = 5$

Now, let's check the conditions :

$r_1 + r_2 = \frac{5}{2} + 5$ = $\frac{15}{2}$

Since $d = \frac{15}{2} = r_1 + r_2$, the circles touch each other externally, and there are 3 common tangents.

Solutions · 1