Geo4.41

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Prove that the tangent to the circle x^{2}+y^{2}=5\, at (1,-2) also touches the circle x^{2}+y^{2}-8x+6y+20=0\, and find the point of contact.

Centre of the given first circle is (0,0) and radius is{\sqrt  {5}}\,

Slope of the line joining the centre and (1,-2) is {\frac  {-2-0}{1-0}}=-2\,

Now the slope of the tangent is 1/2.

The equation of the tangent which has a point (1,2) and slope of -1/2 is

y+2={\frac  {1}{2}}(x-1)\,

2y+4=x-1\,

Hence the equation of the tangent is x-2y-5=0\,</math>

Now the centre of the second circle is (4,-3) and radius is {\sqrt  {16+9-20}}={\sqrt  {5}}\,

The distance from the centre to the line x-2y-5=0 is

{\frac  {4+6-5}{{\sqrt  {5}}}}={\sqrt  {5}}\,

Hence the distance from the centre to the line is equal to the radius,hence it touches the second circle.

Let(x1,y1) be the point of contact.

The distance between centre and the point is equal to radius,

Now

{\sqrt  {(x_{1}-4)^{2}+(y_{1}+3)^{2}}}={\sqrt  {5}}\,

(x_{1}-4)^{2}+(y_{1}+3)^{2}=5\,

The point lies on the tangent,

Hence

x_{1}-2y_{1}-5=0\,

x_{1}=2y_{1}+5\,

Now

(2y_{1}+1)^{2}+(y_{1}+3)^{2}=5\,

4y_{1}^{{2}}+1+4y_{1}+y_{1}^{{2}}+9+6y_{1}=5\,

5y_{1}^{{2}}+10y_{1}+5=0\,

y_{1}^{{2}}+2y_{1}+1=0\,

(y_{1}+1)^{2}=0\,

y_{1}=-1\,

Hence the value of x1=3.

Now the point of contact is (3,-1).


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