# coordinate geometry

#### prasum

show that the locus of a point which moves in such a way that square of its distance from the base of an isosceles triangle is equal to the rectangle under its distance from other sides is a circle.

#### Opalg

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show that the locus of a point which moves in such a way that square of its distance from the base of an isosceles triangle is equal to the rectangle under its distance from other sides is a circle.
I can do this by coordinate geometry, but I don't see a synthetic (euclidean) proof of it.

If the triangle is PQR, with PQ = PR, take P, Q, R to be the points (0,b), (–a,0), (a,0) respectively. The equation of the line PQ is $$\displaystyle bx+ay-ab=0$$ and the equation of PR is $$\displaystyle bx -ay +ab = 0$$.

If I understand the question correctly, you want to find the locus of a point X = (x,y) which moves so that the square of its distance from QR is equal to the product of its distances from PQ and PR.

With the above choices of P, Q and R, the distance from X to PQ is $$\displaystyle \frac{a(b-y)+bx}{\sqrt{a^2+b^2}}$$, and the distance from X to PR is $$\displaystyle \frac{a(b-y)-bx}{\sqrt{a^2+b^2}}$$. The distance from X to QR is |y|. So the locus of X is given by the equation $$\displaystyle \dfrac{a^2(b-y)^2 - b^2x^2}{a^2+b^2} = y^2$$. This simplifies to $$\displaystyle x^2 + \bigl(y+\frac{a^2}b\bigr)^2 = \frac{a^2}{b^2}(a^2+b^2)$$, which represents a circle with centre at $$\displaystyle \bigl(0,-\frac{a^2}b\bigr)$$ and radius $$\displaystyle \frac ab\sqrt{a^2+b^2}$$.

In terms of the geometry of the picture, this circle has its centre C on the perpendicular bisector of QR. It passes through Q and R, and is tangential to PQ and PR.

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#### prasum

thanks man

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