Question

Find the equation to the tangent to the circle ${{\text{x}}^2} + {{\text{y}}^2} = {{\text{a}}^2}$ which
Is parallel to the straight line y = mx + c
Is perpendicular to the straight line y = mx + c
Passes through the point (b, 0)
Make with the axes a triangle whose area is ${{\text{a}}^2}$.

Answer 1

Hint: In order to deduct the tangent equations for these specific cases, we refer to the formulae of tangent equation to a circle equation and pick the appropriate one.

Complete step-by-step answer:
Parallel to the line y = mx + c
We know parallel lines have the same slopes, hence the slope of the equation of the tangent to be found is also m.
We know the tangent to a circle equation ${{\text{x}}^2} + {{\text{y}}^2} = {{\text{a}}^2}$for the line of the form y = mx + c is ${\text{y = mx }} \pm {\text{ a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$, where m is the slope of the line and a is the radius of the circle.
Hence, the tangent equation to the circle parallel to the given line is ${\text{y = mx }} \pm {\text{ a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$.
$\Rightarrow {\text{mx - y}} \pm {\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} = 0$

Perpendicular to the line y = mx + c
We know, perpendicular lines have slopes such that their product is -1, i.e. ${{\text{m}}_1} \times {{\text{m}}_2} = - 1$, where ${{\text{m}}_1}$ and ${{\text{m}}_2}$are the slopes of each line respectively.
The slope of the given line is m, hence the slope of the equation of the tangent to be found is $- \dfrac{1}{{\text{m}}}$.
We know the tangent to a circle equation ${{\text{x}}^2} + {{\text{y}}^2} = {{\text{a}}^2}$for the line of the form y = mx + c is ${\text{y = mx }} \pm {\text{ a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$, where m is the slope of the line and a is the radius of the circle.
Hence, the tangent equation to the circle parallel to the given line is ${\text{y = - }}\dfrac{1}{{\text{m}}}{\text{x }} \pm {\text{ a}}\sqrt {\left[ {1 + {{\left( { - \dfrac{1}{{\text{m}}}} \right)}^2}} \right]}$
$\Rightarrow {\text{y = - }}\dfrac{{\text{x}}}{{\text{m}}} \pm {\text{ a}}\sqrt {\left[ {1 + \dfrac{1}{{{{\text{m}}^2}}}} \right]}$
$\Rightarrow {\text{my = - x}} \pm {\text{a}}\sqrt {\left[ {{{\text{m}}^2} + 1} \right]} \\\ \Rightarrow {\text{x + my }} \mp {\text{a}}\sqrt {\left[ {{{\text{m}}^2} + 1} \right]} = 0 \\\$

Passes through the point (b, 0)
We know the tangent to a circle equation ${{\text{x}}^2} + {{\text{y}}^2} = {{\text{a}}^2}$at a point $\left( {{{\text{x}}_1}{\text{,}}{{\text{y}}_1}} \right)$is given by ${\text{x}}{{\text{x}}_1} + {\text{y}}{{\text{y}}_1} = {{\text{a}}^2}$where a is the radius of the circle.
Hence, the tangent equation to the circle through (b, 0) is ${\text{x}}\left( {\text{b}} \right) + {\text{y}}\left( 0 \right) = {{\text{a}}^2}$
$\Rightarrow {\text{xb = }}{{\text{a}}^2}$.

Makes with the axes a triangle whose area is ${{\text{a}}^2}$
We know the tangent to a circle equation ${{\text{x}}^2} + {{\text{y}}^2} = {{\text{a}}^2}$ for the line of the form y = mx + c is ${\text{y = mx }} \pm {\text{ a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$, where m is the slope of the line and a is the radius of the circle.
${\text{y = mx + a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$
(We consider only the positive sign and divide the entire equation with ${\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$)

$$\Rightarrow \dfrac{{\text{y}}}{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }}{\text{ = }}\dfrac{{{\text{mx}}}}{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }}{\text{ + }}\dfrac{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }}{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }} \\\ \Rightarrow \dfrac{{\text{x}}}{{\dfrac{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }}{{ - {\text{m}}}}}} + \dfrac{{\text{y}}}{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }} = 1{\text{ - - - - }}\left( 1 \right) \\\$$

We know the area formed by the line of the form $\dfrac{{\text{x}}}{{\text{a}}} + \dfrac{{\text{y}}}{{\text{b}}} = 1$ with the axes is $\dfrac{1}{2}|{\text{ab|}}$, where a and b are x and y intercepts respectively.
Comparing this with equation (1) we get a = $\dfrac{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }}{{ - {\text{m}}}}$and b = ${\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$
Hence area of triangle formed = $\dfrac{1}{2} \times \dfrac{{{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} }}{{| - {\text{m|}}}} \times {\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]}$
= $\dfrac{1}{2} \times \dfrac{{{{\left( {{\text{a}}\sqrt {\left[ {1 + {{\text{m}}^2}} \right]} } \right)}^2}}}{{| - {\text{m|}}}}$
= $\dfrac{1}{2} \times \dfrac{{{{\text{a}}^2}\left( {1 + {{\text{m}}^2}} \right)}}{{| - {\text{m|}}}}$
Given area of triangles formed = ${{\text{a}}^2}$
Therefore, ${{\text{a}}^2}$= $\dfrac{1}{2} \times \dfrac{{{{\text{a}}^2}\left( {1 + {{\text{m}}^2}} \right)}}{{| - {\text{m|}}}}$
$\Rightarrow \dfrac{{{\text{a}}\sqrt {1 + {{\text{m}}^2}} }}{{\text{m}}} = \pm {\text{2}}{{\text{a}}^2} \\\ \Rightarrow {{\text{m}}^2} + 1 \pm {\text{2m = 0}} \\\ \Rightarrow {\text{m = - 1 or 1}} \\\$
Hence the equation of the tangent forming an area ${{\text{a}}^2}$with the axes is
${\text{y = }} \pm {\text{x}} \pm {\text{a}}\sqrt {1 + 1} \\\ \Rightarrow {\text{x}} \pm {\text{y = }} \pm {\text{a}}\sqrt 2 \\\$

Note – In order to solve this type of problems the key is to have good knowledge in the concepts of parallel and perpendicular lines. Also the tangent equation to a circle formula with respect to the specific conditions.
Some additional formulae:
The tangent to a circle equation of the form ${{\text{x}}^2} + {{\text{y}}^2} + 2{\text{gx + 2fy + c = 0}}$at point $\left( {{{\text{x}}_1},{{\text{y}}_1}} \right)$is${\text{x}}{{\text{x}}_1}{\text{ + y}}{{\text{y}}_1}{\text{ + g}}\left( {{\text{x + }}{{\text{x}}_1}} \right) + {\text{f}}\left( {{\text{y + }}{{\text{y}}_1}} \right) + {\text{c = 0}}$.
The tangent to a circle equation of the form ${{\text{x}}^2} + {{\text{y}}^2}{\text{ = }}{{\text{a}}^2}$at point $\left( {{\text{acos}}\theta ,{\text{asin}}\theta } \right)$ is ${\text{xcos}}\theta {\text{ + ysin}}\theta {\text{ = a}}$.