The equation of a circle is $y^2 + x^2 = 4$
Two tangents of this circle intersect at point $(10, -5)$, what is the equation of these tangents?
I've tried to work the point of where one of the tangents meets the circle as $(a, b)$ and substitute it into the equation of the circle, and the equation of the tangent – which is of course perpendicular to the equation of the radius that ends at point $(a, b)$ however that proved fruitless as far as I can tell, are there any other methods?
Best Answer
HINT: set as $y=ax+b \, \, $ the equation of one of the tangents. Since you know that the line passes through point $(10,-5)$, you can transform the equation of the line to eliminate $b $ and to keep only the parameter $a $. Then, you have a system to solve to find the intersection points: so you can obtain a second degree equation to be solved for $x $. Take the determinant (it will be a second degree expression containing $a $ ) and find the two values of $a $ for which the determinant is zero. These will be the two slopes of the tangents. From this you can easily get the intercepts as well.