Question

Find the equation of a line tangent to the circle (x−1)^2+y^2=26 at the point (2, −5).

Answer 1

Center of circle is ( 1, 0).

Slope of normal passing though ( 1, 0 ) and ( 2, -5 ) is :

$m_p=\dfrac{0-(-5)}{1-2}\\\\m_p=-5$

So, slope of tangent will be :

$m_t=-\dfrac{1}{m_n}\\\\m_t=\dfrac{1}{5}$

Equation of tangent :

$y-(-5)=\dfrac{1}{5}(x-2)\\\\5y+25=x-2\\\\5y-x+27=0$

Hence, this is the required solution.