If the angle between the tangent from $(0,0)$ to the circle ${x}^{2}+{y}^{2}+10x+10y+40=0$ is ${\tan}^{-1}(m)$, then $m=$
- $\dfrac{3}{4}$
- $\dfrac{1}{2}$
- $2$
- $\dfrac{4}{3}$
The circle is (x+5)^2 + (y+5)^2 = 10. Distance from origin to center is sqrt(50). Tangent length L = sqrt(d^2 - r^2) = sqrt(50 - 10) = sqrt(40). The angle theta between tangents is given by tan(theta/2) = r/L = sqrt(10)/sqrt(40) = 1/2. The question asks for tan(theta) = 2 * tan(theta/2) / (1 - tan^2(theta/2)) = 2 * (1/2) / (1 - 1/4) = 1 / (3/4) = 4/3.
The center of the circle is found using the coefficients of x and y, which gives the point (-5, -5). The radius of the circle is the square root of 25 plus 25 minus 40, which equals root 10. The distance from the origin to the center is root of (-5) squared plus (-5) squared, equaling root 50. Using the property that the angle between tangents is 2 times arctan of radius divided by distance, we get 2 times arctan of root 10 divided by root 50, simplifying to 2 times arctan of 1 over root 2. Using the double angle formula for tangent, tan of 2 times arctan of 1 over root 2 equals 2 times 1 over root 2 divided by 1 minus 1 half, which simplifies exactly to 4 over 3.