The equation of the line that is tangent to the circle at (8, -2) is:
[tex]x=8[/tex]
We know that the line which is tangent at a point on a circle is perpendicular to the line joining the center of the circle and that point.
Here we are given the equation of a circle as:
[tex](x-3)^2+(y+2)^2=25[/tex]
The center of the circle is at: (3,-2)
( Since, the standard form of a circle with center at (h,k) and radius r is given by:
[tex](x-h)^2+(y-k)^2=r^2[/tex] )
Also, the equation of a line joining (3.-2) and (8,-2) is given by:
[tex]y-(-2)=\dfrac{-2-(-2)}{8-3}\times (x-3)\\\\i.e.\\\\y+2=\dfrac{-2+2}{5}\times (x-3)\\\\i.e.\\\\y+2=\dfrac{0}{5}\times (x-3)\\\\i.e.\ y+2=0\\\\i.e.\ y=-2[/tex]
Also, we know that the slope of this line is zero.
Also, we know that if two lines are perpendicular with slope m and m' respectively then,
[tex]m\cdot m'=-1\\\\i.e.\\\\m'=\dfrac{-1}{m}[/tex]
Hence, we get that the slope of the tangent line is:
[tex]\dfrac{-1}{0}[/tex]
Also, we know that:
The equation of a line with given slope m' and a passing through point (a,b) is given by:
[tex]y-b=m'(x-a)[/tex]
Here (a,b)=(8,-2)
and [tex]m'=\dfrac{-1}{0}[/tex]
i.e. the equation of the tangent line is:
[tex]y-(-2)=\dfrac{-1}{0}(x-8)\\\\i.e.\\\\y+2\times 0=-1(x-8)\\\\i.e.\\\\-1\times (x-8)=0\\\\i.e.\\\\(x-8)=0\\\\i.e.\\\\x=8[/tex]
