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The common tangents of the circles

$\mathrm{x^2+y^2+14 x-4 y+28=0 \text { and } x^2+y^2-14 x+4 y-28=0 \text { are }}$ ;
Option: 1
$\mathrm{y=7}$

Option: 2
$\mathrm{28 x+45 y+371=0}$

Option: 3
$\mathrm{x+2=0}$

Option: 4
$\mathrm{45 x-28 y+106=0}$

Answers (1)

best_answer

$\mathrm{{A}(-7,2), \sqrt{49+4-28}=5=r_1}$

$\mathrm{B(7,-2), 9=r_2}$

$\mathrm{\mathrm{AB}>r_1+r_2 \Rightarrow}$ circles are non-intersecting

Q is $\mathrm{\left(\frac{-63+35}{9+5}, \frac{18-10}{9+5}\right)=Q\left(-2, \frac{4}{7}\right)}$

P is $\mathrm{\left(\frac{-63-35}{9-5}, \frac{18+10}{9-5}\right)=\left(-\frac{49}{2}, 7\right)}$

$\therefore$ The direct common tangents are

$\mathrm{y-7=M\left(x+\frac{49}{2}\right)}$

On applying the condition of tangency to circle with centre A (–7, 2) and radius 5, we obtain

$\mathrm{\frac{2-7-M\left(-7+\frac{49}{2}\right)}{\sqrt{1+M^2}}=5 \Rightarrow M=0,-\frac{28}{45}}$

Direct common tangents are $\mathrm{y-7=0\: \& \: y-7=-\frac{28}{45}\left(x+\frac{49}{2}\right)}$

$\mathrm{28 x+45 y+371=0}$

Transverse common tangents are given by

$\mathrm{y-\frac{4}{7}=m(x+2)}$

On applying the condition of tangency to the circle with centre (–7, 2) and radius 5, we obtain

$\mathrm{\begin{aligned} & \frac{2-\frac{4}{7}-m(-7+2)}{\sqrt{1+m^2}}=5 \\ & \frac{10}{7}+5 m=5 \sqrt{1+m^2} \end{aligned}}$

Or $\mathrm{\left(\frac{2}{7}+m\right)^2=1+m^2}$

Or $\mathrm{\begin{aligned} & m^2(1-1)+\frac{4}{7} m+\frac{4}{49}-1=0 \\ & m=\frac{45}{49} \times \frac{7}{4}=\frac{45}{7 \times 4}=\frac{45}{28} \end{aligned}}$

And, the 2nd value of m is $\mathrm{\infty}$ .
The transverse common tangents are

$\mathrm{\begin{gathered} y-\frac{4}{7}=\frac{45}{28}(x+2) \text { or } \frac{45}{28} x-y+\frac{45}{19} \times \frac{4}{7}=0 \\ \text { or } \frac{45}{28} x-y+\frac{53}{14}=0 \end{gathered}}$

Or $\mathrm{45 x-28 y+106=0}$

And, $\mathrm{x+2=0}$

Posted by

Ajit Kumar Dubey

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