Find the equation to the common chord of the two circles x^2+y^2-4x-10y-7=0 and 2x^2+2y^2-5x+3y+2=0 and show that this chord is perpendicular to line joining the centres of the two circles?

1 Answer
Feb 13, 2018

#"The eqn. of common chord is "3x+23y+16=0#.

Kindly refer to the Explanation for the complete Solution.

Explanation:

Let the given circles be,

# S_1 : x^2+y^2-4x-10y-7=0, and,#

# S'_2 : 2x^2+2y^2-5x+3y+2=0#.

We rewrite #S'_2" as "S_2 : x^2+y^2-5/2x+3/2y=-1#.

Then, completing the square, #S_2# becomes,

# S_2 : (x-5/4)^2+(y+3/4)^2=(3sqrt2/4)^2#.

Similarly, # S_1 : (x-2)^2+(y-5)^2=(6)^2#.

We find that the Centres #C_i" of "S_i (i=1,2)# are,

#C_1=C_1(2,5), and, C_2=C_2(5/4,-3/4)#.

If #P(X,Y)# is any arbitrary point on the reqd. common

chord, say #l#, then #P in S_1, and, P in S_2#.

#P in S_1 rArr X^2+Y^2-4X-10Y-7=0......(ast^1)#.

#P in S_2 rArr X^2+Y^2-5/2X+3/2Y+1=0......(ast^2)#.

#:. (ast^2)-(ast^1) rArr 3/2X+23/2Y+8=0, or,#

# 3X+23Y+16=0...........................................(ast)#.

We note that #(ast)# is a linear eqn. in #RR^2#, and,

therefore, represents a line passing through the points of

intersection of #S_1 and S_2,# their common chord #l#.

Switching from #(X,Y)# to the conventional #(x,y)#,

#:. l : 3x+23y+16=0#.

Next, to show that #l bot C_1C_2#.

We have, # l :y=-3/23x -16/3= #, so its slope #m_1=-3/23#.

The slope #m_2" of "C_1C_2" is "{5-(-3/4)}/(2-5/4)=23/3#.

#because m_1*m_2=-1, :., l bot C_1C_2#.

Hence, the Proof.