Geo4.10

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Find the power of the point (a+b,a-b) with respect to x^{2}+y^{2}+2bx-3b^{2}=0\,


Let the point be P and centre be C and radius r

Power of the point is

[CP]^{2}-r^{2}\,

Centre of the given circle is

\left({\frac  {-2b}{2}},{\frac  {0}{2}}\right)\,

\left(-b,0\right)\,

Now

[CP]^{2}=(a+b+b)^{2}+(a-b-0)^{2}\,

[CP]^{2}=(a+2b)^{2}+(a-b)^{2}\,

r^{2}=b^{2}+3b^{2}=4b^{2}\,

Hence the power of the point is

[CP]^{2}-r^{2}=(a+2b)^{2}+(a-b)^{2}-4b^{2}\,

a^{2}+4b^{2}+4ab+a^{2}+b^{2}-2ab-4b^{2}\,

2a^{2}+2ab+b^{2}\,


Main Page:Geometry:Circles