Equation of the tangent to the circle, at the point (1,-1), whose centre is the point of i…

Question

Equation of the tangent to the circle, at the point $(1,-1)$, whose centre is the point of intersection of the straight lines $x-y=1$ and $2x+y=3$, is:

Accepted Answer

Point of intersection of lines

x - y = 1   and  2x + y = 3 is $$\left( {{4 \over 3},{1 \over 3}} \right)$$

Slope of OP = $${{{1 \over 3} + 1} \over {{4 \over 3} - 1}}$$ = $${{{4 \over 3}} \over {{1 \over 3}}}$$ = 4

Slope of tangent = - $${1 \over 4}$$

Equation of tangent    y + 1 = - $${1 \over 4}$$ (x - 1)

4y + 4 = - x + 1

x + 4y + 3 = 0

Solutions · 1