If a line y=mx+c is a tangent to the circle (x-3)^2+y^2=1 and it is perpendicular to a lin…

Question

If a line $y=mx+c$ is a tangent to the circle $(x-3)^2+y^2=1$ and it is perpendicular to a line $L_1$, where $L_1$ is the tangent to the circle $x^2+y^2=1$ at the point $\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$, then:

Accepted Answer

For circle x2 + y2
 = 1 tangnet at point P$$\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$ is

T = 0

$$ \Rightarrow $$ $${1 \over {\sqrt 2 }}x + {1 \over {\sqrt 2 }}y - 1 = 0$$

$$ \Rightarrow $$ x + y - $$\sqrt 2 $$ = 0

Perpendicular to the line is

x - y + c = 0

This is tangent to the circle (x -- 3)2
 + y2
 = 1

Also we know, perpendicular distance from center = radius

$$ \therefore $$ $$\left| {{{3 - 0 + c} \over {\sqrt 2 }}} \right|$$ = 1

$$ \Rightarrow $$ |3 + c| = $${\sqrt 2 }$$

$$ \Rightarrow $$ $${\left( {3 + c} \right)^2} = 2$$

$$ \Rightarrow $$ 9 + c2
 + 6c = 2

$$ \Rightarrow $$ c2
 + 6c + 7 = 0

Solutions · 1