#(x-7)² + (y+24)² = 196# Min# (x²+y²)=# ?

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Answer 1 · 2017-05-02T12:45:32

See below.

#(x_0,y_0) = (77/25, -264/25)#

This is the tangency point between the circle

#x^2+y^2=11^2#

and the circle

#(x-7)^2+(y+24)^2=14^2#

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Answer 2 · 2017-05-02T13:38:28

#121.#

Observe that, # S : (x-7)^2+(y+24)^2=196,# represents a circle

with Centre at #C( 7,-24)# and Radius #r=sqrt196=14.#

The Parametric Eqns. of #S# are given by,

#x=7+14costheta, y=-24+14sintheta, theta in [0,2pi).#

#:. x^2+y^2=(7+14costheta)^2+(-24+14sintheta)^2,#

#=(7^2+2*7*14costheta+14^2cos^2theta)#

#+(24^2-2*24*14sintheta+14^2sin^2theta),#

#=(7^2+24^2)+2*14(7costheta-24sintheta)+14^2(1),#

#=25^2+14^2+28*25(7/25costheta-24/25sintheta).#

Letting, #7/25=cosalpha," so that, "sinalpha=24/25,# we find,

#x^2+y^2=25^2+14^2+25*28cos(theta+alpha).#

Since, the Minimum Value of #cos" function is, "-1,#

We conclude that the Reqd. Minima is, # 25^2+14^2-25*28#

#=25^2-2*25*14+14^2=(25-14)^2=121.#

Enjoy Maths.!