Question

Two chords of length 5 are drawn from any point (3,4) on the circle $4x^2+4y^2-24x-7y=0$ then their equations are given by $y-4=\pm \frac{4}{3}(x-3)$

• A. True
• B. False

Answer 1

The correct option is A True

Let the chord be $y-4=m(x-3)$

or $mx-y+(4-3m)=0$ .............(1)

$AP=AQ=5$

$\therefore 5=2\sqrt{(r^2-p^2)}$

Circle is $x^2+y^2-6x-\frac{7}{4}y=0$

C is $(3,\frac{7}{8})andr=\frac{25}{8}$

$3m-\frac{7}{8}+4-3m$

$\therefore p=\frac{3m-\frac{7}{8}+4-3m}{\sqrt{1+m^2}}=\frac{25}{8\sqrt{1+m^2}}$

Hence from (1), we get

$\frac{25}{4}={\left(\frac{25}{8}\right)}^2-{\left(\frac{25}{8}\right)}^2\cdot \frac{1}{(1+m^2)}$

or $16(m^2+1)=25m^2\therefore m=\pm 4/3$

Hence from (1) the lines are $4x-3y=0$ and $4x+3y-24=0$.

Alternative method :

On dividing by 4,the circle is

$(x-3{)}^2+{(y-\frac{7}{8})}^2={\left(\frac{25}{8}\right)}^2$ ............(1)

Any line through $(3,4)$ is

$y-4=tan\theta (x-3)$ .........(2)

or $\frac{x-3}{cos\theta }=\frac{y-4}{sin\theta }=r=5forPorQ$

$(5cos\theta +3,5sin\theta +4)$ is the other extremity of chord which will lie on the circle (1).

$\therefore (5cos\theta {)}^2+{(5sin\theta +\frac{25}{8})}^2={\left(\frac{25}{8}\right)}^2$

or $25+\frac{125}{4}sin\theta =0$

This gives $sin\theta =-\frac{4}{5}\therefore cos\theta =\pm \frac{3}{5}.$

Hence $tan\theta =\pm \frac{4}{3}=m$

Putting for $tan\theta$ in (2), the required lines are $4x=3y$ and $4x+3y-24=0.$