"If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2 + y2 = 1 at the point $\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$, then :"

Question

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 "

If a line, y = mx + c is a tangent to the circle, (x – 3)² + y² = 1 and it is perpendicular to a line L₁, where L₁ is the tangent to the circle, x² + y² = 1 at the point $\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$, then :

Solution

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If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2 + y2 = 1 at the point $\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$, then :

Solution

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More Circles questions

Drill 25 more like these. Every day. Free.

If a line, y = mx + c is a tangent to the circle, (x – 3)² + y² = 1 and it is perpendicular to a line L₁, where L₁ is the tangent to the circle, x² + y² = 1 at the point $\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$, then :