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  •  Find the equations of the circles which pass through the points A(–1, 4), B(1, 2) and which touch the line 3x – y – 3 = 0.  <stron

 Find the equations of the circles which pass through the points A(–1, 4), B(1, 2) and which touch the line 3x – y – 3 = 0.

 

Option: 1

x^2+y^2+2 x-y-1=0 \: \&\: x^2+y^2+4 x-7 y-5=0


Option: 2

x^2+y^2+4 x-2 y-5=0 \: \& \: x^2+y^2-x-7 y+10=0


Option: 3

4 x^2+4 y^2+9 x-3 y-5=0 \: \&\: x^2+y^2-x+7 y-6=0


Option: 4

None of these


Answers (1)

best_answer

Let C(h, k) be the centre of a circle which satisfies the required conditions. Consider M(0, 3) the mid-point of the line AB. Then CM perpendicular AB

\begin{aligned} &\therefore \frac{\mathrm{k}-3}{\mathrm{~h}-0} \times \frac{4-2}{-1-1}=-1\\ &\therefore \mathrm{h}=\mathrm{k}-3 . \end{aligned} -------------(i)

Now length of perpendicular from C(h, k) on 3x – y – 3 = 0 is equal to CA. 


\therefore\left|\frac{3 \mathrm{~h}-\mathrm{k}-3}{\sqrt{10}}\right|=\sqrt{(\mathrm{h}+1)^2+(\mathrm{k}-4)^2} \text {. }

\mathrm{\begin{aligned} & \text { Square both sides } \quad \therefore(3 h-k-3)^2=10\left[(h+1)^2+(k-4)^2\right] \\ & \text { Or } 9 h^2+k^2-6 h k-18 h+6 k+9=10\left[h^2+k^2-8 k+17\right] \\ & \text { or } h^2+9 k^2+6 h k+38 h-86 k+161=0 . \\ & \therefore(k-3)^2+9 k^2+6(k-3) k+38(k-3)-86 k+161=0 \quad \text { or } 16 k^2-72 k+56=0 \text { or } 2 k^2-9 k+7=0 \\ & \therefore(k-1)(2 k-7)=0 \quad \therefore k=1 \text { or } 7 / 2 \text { and } h=-2 \text { or } 1 / 2 \end{aligned}}

Hence the 2 required circles are with centres (–2, 1) and (1/2, 7/2) and \mathrm{\operatorname{radii} \sqrt{10} \text { and } \sqrt{5 / 2}} respectively.

Their equations are \mathrm{(x+2)^2+(y-1)^2=10 \text { and }(x-1 / 2)^2+(y-7 / 2)^2=5 / 2}

\mathrm{\text { or } x^2+y^2+4 x-2 y-5=0 \text { and } x^2+y^2-x-7 y+10=0}

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