Question

Angle between pair of tangents drawn from point $(4,3)$ to the circle $x^2+y^2-4x+6y-3=0$ is:

• A. ${\tan }^{-1}2\sqrt{(3}$
• B. ${\tan }^{-1}2\sqrt{6}$
• C. ${\tan }^{-1}2\sqrt{2}$
• D. ${\tan }^{-1}2\sqrt{6}$
• E. ${\tan }^{-1}3\sqrt{2}$

Answer 1

The correct option is A ${\tan }^{-1}2\sqrt{(3}$

$x^2+y^2-4x+6y-3=0-0$

$\text{Eq of a circle with centre}(2,-3)$

$\text{Equation of tangents to the given circle is}$

$(y-k)=m(x-h)\pm a\sqrt{1+m^2}$

a is the radius of the given circle $h$ and $k$ are coordinates of centre of circle

$a=\sqrt{(2{)}^2+(3{)}^2-(-3)}=4$

$y+3$ = $m(x-2)\pm$4

$\sqrt{1+m^2}$

tangents are passing through (4,3)

$7=m\pm 4\sqrt{1+m^2}$

$7-m=\pm 4\sqrt{1+m^2}$

Squaring on both sides

$(7-m{)}^2=16(1+m^2)$

$m^2+49-14m=16m^2+16$

Angle between the tangents is

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$m$ and $m_2$ are the slope of the tangents and they are the roots of eq

$\tan \theta =\mid c\sqrt{2176}$ $\frac{15}{1-\frac{33}{15}}\mid \}$ $=$ $-\sqrt{2176}$/28

$\tan \theta =$ $\frac{2\sqrt{34}}{7}$

$\theta ={\tan }^{-1}\left(\frac{2\sqrt{34}}{7}\right)$