Geo4.39

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Show that the line x+y+1=0 touches the circle x^{2}+y^{2}-3x+7y+14=0\, and find the point of contact.

The centre of the given circle is (3/2,-7/2) and radius is {\sqrt  {{\frac  {9}{4}}+{\frac  {49}{4}}-14}}\,

Radius is {\frac  {1}{{\sqrt  {2}}}}\,

The distance from the centre of the circle to the given line is

|{\frac  {{\frac  {3}{2}}-{\frac  {7}{2}}+1}{{\sqrt  {2}}}}|={\frac  {1}{{\sqrt  {2}}}}\,

Now the perpendicular distance from the centre of the circle to the given line is equal to the radius of the circle. Hence the line touches the circle.

Let the point of contact be (x1,x2)

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

{\sqrt  {(x_{1}-{\frac  {3}{2}})^{2}+(y_{1}+{\frac  {7}{2}})^{2}}}={\frac  {1}{{\sqrt  {2}}}}\,

2[(x_{1}-{\frac  {3}{2}})^{2}+(y_{1}+{\frac  {7}{2}})^{2}]=1\,

The point lies on the given line,hence

x_{1}+y_{1}+1=0\,

x_{1}=-(y_{1}+1)\,

Substituting this value in the above equation,we get

2[(-y_{1}-{\frac  {5}{2}})^{2}+(y_{1}+{\frac  {7}{2}})^{2}]=1\,

2[y_{1}^{{2}}+{\frac  {25}{4}}+5y_{1}+y_{1}^{{2}}+{\frac  {49}{4}}+7y_{1}]=1\,

4y_{1}^{{2}}+24y_{1}+36=0\,

y_{1}^{{2}}+6y_{1}+9=0\,

(y_{1}+3)^{2}=0\,

y_{1}=-3\,

Now the value of x1 is 2

Hence the point of contact is (2,-3)


Main Page:Geometry:Circles