Question

Find the equation to the tangent at a to the circle (x-a)^2+(y-b)^2=r^2

Answer 1

Equation is - $(x - a)cos\alpha + (y - b)sin \alpha = \pm r$

Step-by-step explanation:

From the figure, it is quite clear that two such tangents are possible  

only if they are parallel to each other

Therefore, the Slope of the tangent $(m_1)$ is perpendicular to slope of radius $(m_2)$

$m_1.m_2 = -1 (m_1 = tan\alpha)$

$tan\alpha (m_2) = -1$

So, $m_2 = - cot\alpha$

Hence, using standard result equation of tangents of slope m is given by -

$(y - b) = m(x - a) \pm r\sqrt (1 + m^2)$

$(y - b) = - cot \alpha (x - a) \pm r\sqrt (1 + cot^2\alpha)$

$(y - b) = -\frac {cos \alpha}{sin \alpha}(x - a) \pm rcosec \alpha$

$sin(y - b) = - cos\alpha (x - a) \pm rcosec \alpha sin \alpha\ (As\ sin\alphacosec\alpha =1)$

Hence, $(x - a)cos\alpha + (y - b)sin \alpha = \pm r$