Geo4.27

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Find the equation to the circle passing through the points (1,-2),(4,-3) and having the centre on the line 3x+4y=7\,

Let the equation of the circle be

x^{2}+y^{2}+2gx+2fy+c=0\,

The circle passes through (1.-2)

Hence

1^{2}+(-2)^{2}+2g(1)+2f(-2)+c=0\,

2g-4f+c=-5\,

Let this equation be 1.

The circle passes through (4.-3)

Hence

4^{2}+(-3)^{2}+2g(4)+2f(-3)+c=0\,

8g-6f+c=-25\,

Let this equation be 2.

Now the centre (-g,-f)lies on 3x+4y=7.

Hence

-3g-4f=7\,

3g+4f=-7\,

Let this equation be 3.

Solving 1 and 3 we get

5g+c=-12\,

Let this be 4.

Now solving 2 and 3 we get

50g+4c=-142\,

Let this be 5.

Solving 4 and 5 we get

-30g=94,g={\frac  {-94}{30}}\,

Now

5\cdot {\frac  {-94}{30}}+c=-12\,

c={\frac  {110}{30}}\,

The value of f is

3({\frac  {-94}{30}}+4f=-7\,

4f={\frac  {72}{30}}\,

Hence the equation of the circle is

15(x^{2}+y^{2})-94x-18y+55=0\,


Main Page:Geometry:Circles