Show that the circles x^2+y^2+6(x-y)+9=0 touches the co-ordinate axes.Also find the equation of the circle which passes through the common point of intersection of the above circle and the straight line x-y+4=0 and which also passes through the origin?

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Answer 1 · 2018-02-28T04:47:41

# 4x^2+4y^2+15x-15y=0#.

Let us rewrite the given eqn. of the circle #S# as,

#S : x^2+y^2+6(x-y)+9=0#.

#:. S : x^2+6x+9+y^2-6y+9=9, i.e., #

# (x+3)^2+(y-3)^2=3^2," so that, "# the centre #C# and

radius #r# of #S# are, #C(-3,3), and r=3#.

We know, from Geometry that, the #bot-# distance from

the centre to the tangent equals the radius of the circle.

Now, the #bot-"dist. of "C" to the "X-"Axis",#

#=|y-"co-ordinate of "C|=|3|=3=r, and #,

the #bot-"dist. of "C" to the "Y-"Axis",#

#=|x-"co-ordinate of "C|=|-3|=3=r#.

These prove that #S# touches both the Axes.

Let us denote by #l# the given line # x-y+4=0#.

We seek for a circle #s# that passes through the points of

intersection of #S and l#.

Knowing that, #s# can be represented by, # s : S+lambdal=0, lambda in RR#,

suppose that, #s : x^2+y^2+6x-6y+9+lambda(x-y+4)=0#.

Given that #O(0,0) in s# helps us determine #lambda :#

#O in s rArr 0^2+0^2+6(0)-6(0)+9+lambda(0-0+4)=0#.

#:. lambda=-9/4#.

Hence, #s : 4(x^2+y^2+6x-6y+9)-9(x-y+4)=0, i.e., #

# s : 4x^2+4y^2+15x-15y=0#.