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  • Find the centres of the circles with radius 15 and touching the circle at the point <img a

Find the centres of the circles with radius 15 and touching the circle x^2+y^2=100 at the point (6,-8).

Option: 1

\mathrm{(-3,4) \text { and }(15,-20)}


Option: 2

\mathrm{(3,-4) \text { and }(-15,20)}


Option: 3

\mathrm{(-3,15) \text { and }(4,-20)}


Option: 4

\mathrm{(-3,20) \text { and }(4,15)}


Answers (1)

best_answer

The centre of the circle x^2+y^2=100 is the point (0,0).

\therefore  the centres of the required circles must lie on the line joining (0,0) and (6,-8).

The equation of this line is 4 x+3 y=0.

Using (\mathrm{h}, \mathrm{k}) for the centre, we have 4 h+3 k=0                      ....(i)

and (h-6) 2+(k+8) 2=225                                           ....(ii)

Solving the equations (i) and (ii), the coordinates of the two centres are (-3,4) and (15,-20).

The equations of the two circles are (x+3) 2+(y-4) 2=225

and (x-15) 2+(y+20) 2=225.

Alternatively,
The tangent at (6,-8) to x^2+y^2-100=0 is 3 x-4 y-50=0\left(x x_1+y y_1-r^2=0\right).

The required circle is a member of the system

x^2+y^2-100+k(3 x-4 y-50)=0                               ...(i)

The (radius) ^2 of this circle is \frac{9 k^2}{4}+4 k^2+100+50 kand this is equal to 225 for the required circle.

\begin{aligned} & \therefore 25 k^2+200 k-500=0 \\\\ & k^2+8 k-20=0 \Rightarrow(k+10)(k-2)=0 \end{aligned}                   From \, \, \, (i),

For k=-10, first circle is x^2+y^2-30 x+40 y+400=0.

For k=+2, second circle is x^2+y^2+6 x-8 y-200=0.

Posted by

Suraj Bhandari

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