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=(0-(-5))/(1-2)\\\\m_p=-5`

So, slope of tangent will be :

`m_t=-(1)/(m_n)\\\\m_t=(1)/(5)`

Equation of tangent :

`y-(-5)=(1)/(5)(x-2)\\\\5y+25=x-2\\\\5y-x+27=0`

Hence, this is the required solution.