Question

The angle between the circles $x^2+y^2−4x−6y−3=0$, $x^2+y^2+8x−4y+11=0$

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (C)

$\frac{\pi}{3}$

Answer 2

The equation $x^2+y^2−4x−6y−3=0$ corresponds to a circle with radius $r_1$​, which can be calculated as $\sqrt{2^2+3^2+3^2​}=\sqrt{16}=4$. The center of this circle is (2,3).

The equation $x^2+y^2+8x−4y+11=0$ represents a circle with radius $r_2​$, determined by $\sqrt{4^2+2^2−11^2}​=\sqrt{3}$​. Its center is (−4,2).

To find the angle $\theta$ between these two circles, we can use the cosine formula:
$\text{cos}(180−\theta)=\frac{2r_1​r_2​}{r_1^2​+r_2^2​−(c_1​c_2​)^2}​$

Putting in the values:
$\text{cos}(180−\theta)=\frac{2⋅4⋅3​}{4^2+3^2−(2⋅3)(−4⋅2)}$
$\text{cos}(180−\theta)=\frac{24}{25+37}=\frac{24}{62}​=\frac{12}{31}​$

Now, to find $\theta$:
$\text{cos}(\theta)=\frac{1}{2}$​
$\theta=\frac{\pi}{3}$​

Hence, the angle between these two circles is $\frac{\pi}{3}$ or 60 degrees.