circles

ayushi jain

the tangants x²+ y² = a² having inclinations and intersect at P. If + = 0 .Then the locus of P is

hemang

are the tangents to the circle x^2 + y^2 = a^2??

if yes then reply..

ayushi jain

yeah..

ayushi jain

yeah..

Krishna

i guess it would be a hyperbola with eccentricity root 2.please confirm from some1 and let me know too.

Manasi

cot @ + cot # = 0

=> cot @ = -cot #

=> tan @ = -tan # = m (let) ....................(A)

now Let the point of intersection be P(h,k)

Thus, the equation of the tangents will be

(y-k) / (x-h) = m and, (y-k) / (x-h) = -m

=> y=mx + (k-mh) and y = -mx + (k+mh)

The distance of origin, which is the centre of the circle should be same for the two tangents, and equal to the radius 'a'

Thus (k-mh)/ [(1+m²)]^(1/2) = a = (k+mh)/ [(1+m²)]^(1/2)

=> k-mh = k+mh

=> h =0

The locus is x=0 ...i.e. the y-axis

Also, from the equation (A) ...we can infer, that the inclination angles are supplementary, which is possible only when the point P always lie on y-axis ...

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