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

Question

asked Feb 12, 2021 157k views

1 Answer

DdoGas · Answer 1 · 2021-02-19T04:29:39+0000

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.