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  • Two circles each of radius 5 units touch each other at (1,2). If the equation of their common tangent is 4x + 3y = 10. Find the equation of the two circles.  <div class='qna

Two circles each of radius 5 units touch each other at (1,2). If the equation of their common tangent is 4x + 3y = 10. Find the equation of the two circles.

 

Option: 1

\mathrm{(x+5)^2+(y+5)^2=25 \: and\: (x+3)^2+(y-1)^2=25}


Option: 2

\mathrm{(x-5)^2+(y+5)^2=25\: and\: (x-3)^2+(y+1)^2=25}


Option: 3

\mathrm{(x-5)^2+(y-5)^2=25 \: and \: (x+3)^2+(y+1)^2=25}


Option: 4

None of these


Answers (1)

best_answer

The radius of the required circle is 5 units.

The two circles touch each other externally, at the mid point of the line joining their centres.

Also we know that the common tangents 4x + 3y = 10 and the line joining the centres are perpendicular to each other.

Hence slope of  AB is \mathrm{\tan \theta=3 / 4} 

(where AB is the line joining the centres)
Mid point of AB is (1, 2)

Hence the two centres A and B are obtained by using the parametric form.

\mathrm{\begin{aligned} & \frac{x-1}{\cos \theta}=\frac{y-2}{\sin \theta}= \pm r \\ & \frac{x-1}{(4 / 5)}=\frac{y-2}{(3 / 5)}= \pm 5 \\ & x=\frac{4}{5}( \pm 5)+1 \\ & y=\frac{3}{5}( \pm 5)+2 \end{aligned}}

Hence the centres of the required circles are  \mathrm{\mathrm{C}_1 \equiv(5,5) \text { and } \mathrm{C}_2 \equiv(-3,-1)}

We get the equations of the circles as \mathrm{(x-5)^2+(y-5)^2=25 \text { and }(x+3)^2+(y+1)^2=25}

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