Question

Find the equation to the tangent to the circle $x^2+y^2=a^2$ which
(i) is parallel to the straight line $y=mx+c$,
(ii) is perpendicular to the straight line $y=mx+c$,
(iii) passes through the point $(b,0)$,
and (iv) makes with the axes a triangle whose area is a^{2}$.

Answer 1

(i) Let the straight line be $y=mx+c_1$

Since it is tangent to the circle $x^2+y^2=a^2$

So, $a=\frac{|0+0-c_1|}{\sqrt{1+m^2}}$

$c_1=a\sqrt{(1+m^2)}$ or, $c_1=-a\sqrt{(1+m^2)}$

Hence, the tangents are

$y=mx+a\sqrt{(1+m^2)}$ and $y=mx-a\sqrt{(1+m^2)}$

(ii) Let the straight line be $y=m_{1}x+c_2$

$m_1=-\frac{1}{m}$

Since it is tangent to the circle $x^2+y^2=a^2$

So, $a=\frac{|0+0-c_2|}{\sqrt{1+\frac{1}{m^2}}}$

$c_2=\frac{a\left(\sqrt{1+m^2}\right)}{m}$ or $c_2=-\frac{a\left(\sqrt{1+m^2}\right)}{m}$

So, the equation of tangent are

$y=-\frac{1}{m}x+\frac{a\left(\sqrt{1+m^2}\right)}{m}$ and $y=-\frac{1}{m}x-\frac{a\left(\sqrt{1+m^2}\right)}{m}$

(iii) Let the straight line be $y=mx+c$

The point passes through $(b,0)$

So, $0=bm+c$

$c=-bm$

Since it is the tangent to the circle $x^2+y^2=a^2$

So,$a=\frac{|0+0+bm|}{1+m^2}$

Solving we get the tangent as

$y=\sqrt{\frac{a^2}{b^2-a^2}}x-bm$ and $y=-\sqrt{\frac{a^2}{b^2-a^2}}x-bm$

(iv) Let the straight line be $y=mx+c$

It cuts $x$-axis at $(-\frac{c}{m},0)$ and $y$-axis at $(0,c)$.

Area of triangle is $a^2$ which is also equal to $\frac{1}{2}.\frac{c}{m}.c$.

$\Rightarrow \frac{c^2}{2m}=a^2$

Since it is the tangent to the circle $x^2+y^2=a^2$

So, $a=\frac{|0+0-c|}{\sqrt{1+m^2}}$

On solving we get,

$m=1$ and $c=\sqrt{2}a$

So, the tangent is, $y=x+\sqrt{2}a$.