Question

Find the equation of a tangent to the circle x2

+y2

-

8x+6y+4=0 at the point (2,-1)​

Answer 1

Equation of the tangent to the circle

$\quad x^{2}+y^{2}+2gx+2fy+c=0$

at the point $(x_{1},y_{1})$ is given by

$\quad xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0$

To find: ( correcting the question )

The equation of a tangent to the circle

$\quad x^{2}+y^{2}-8x+6y+17=0$ at the point $(2,-1)$

Step-by-step explanation:

• The given circle can be written in the form,

• $\quad (x)(x)+(y)(y)+(-4)(2x)+(3)(2y)+17=0$

• Using the above formula, the required tangent to this circle at the point $(2,-1)$, we obtain:

• $\quad (x)(2)+(y)(-1)+(-4)(x+2)+(3)(y-1)+17=0$

• $\Rightarrow 2x-y-4x-8+3y-3+17=0$

• $\Rightarrow -2x+2y+6=0$

• $\Rightarrow x-y-3=0$

• $\Rightarrow x=y+3$

Answer: The required tangent is

$\quad\quad\quad x=y+3$.