In: Math

. Let two circles C1 and C2 intersect at X and Y. Prove that a point P is on the line XY if and only if the power of P with respect to C1 is equal to the power of P with respect to C

The *power of a point theorem*, due to Jakob Steiner,
states that for any line through *A* intersecting a circle
*c* in points *P* and *Q*, the power of the
point with respect to the circle *c* is given up to a sign
by the product

AP.AQ

of the lengths of the segments from *A* to *P* and
*A* to *Q*, with a positive sign if *A* is
outside the circle and a negative sign otherwise: if *A* is
on the circle, the product is zero. In the limiting case, when the
line is tangent to the circle, *P* = *Q*, and the
result is immediate from the Pythagorean theorem.

**Answer** **to** **the**
**question**

Power of p with respect to c1
=(**PO**_{1})²-r²

=PX.PY ( By the definition of

Power of the point)

Power of p with respect to c2= (**PO**_{2})²-r²

=PX.PY

PO = distance between point p and center of circle.

r = radius of circle

Therefore power of P w.r.t to c1= power of P w.r.t to c2

That implies a point P is on the line XY if and only if the power of P with respect to C1 is equal to the power of P with respect to C.

For your
understanding of power of a point

Suppose on a plane we have a point P and a circle (O). Draw a line
through P which intersects with the circle at two points U and V.
Then the value of

PU×PV

is *independent of the choice of the line* PUV.

This means that if we draw another line through P which cuts the
circle at two other points A and B then

**PA**×**PB**=**PU**×**PV**.

This constant value is called *the power of the point* P
*with respect to the circle* (O).

To prove that PU×PV is *independent* of the position of the
line, we will use similar triangles. We will consider two separate
cases: when the point P is *outside* the circle and when the
point P is *inside* the circle.

When P is outside the circle (O)

We consider two triangles PUB and PAV. These two triangles are
similar triangles because they have two pairs of equal angles.
Thus,

PU/PA=PB/PV

From here we obtain the required equality

PA×PB=PU×PV.

When P is inside the circle (O)

When P is inside the circle, we also consider the two triangles PUB and PAV. These two triangles are similar triangles because they have equal angles. So

PU/PA=PB/PV.

This implies the constancy of the power of the point P:

PA×PB=PU×PV

Also, another definition way to calculate

Power of the point = PO^{2} - r^{2}

^{where PO is the distance between the point p and the centre
of circle o, and r is the radius of the circle}

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