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Question 15 of 37

Q.Find the equation and length of the common chord of the two circles : x2+y2+3x+5y+4=0x^2 + y^2 + 3x + 5y + 4 = 0 and x2+y2+5x+3y+4=0x^2 + y^2 + 5x + 3y + 4 = 0.

Telangana TsbieTelangana Board of Intermediate Education 2018Subjective· 4mImportance★★★★★
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The common chord of two circles S1=0S_1=0, S2=0S_2=0 is the line S1−S2=0S_1-S_2=0; then use the perpendicular distance from a centre to find the chord length.

S1≡x2+y2+3x+5y+4=0S_1\equiv x^2+y^2+3x+5y+4=0, S2≡x2+y2+5x+3y+4=0S_2\equiv x^2+y^2+5x+3y+4=0.

Common chord: S1−S2=0⇒(3x+5y)−(5x+3y)=0⇒−2x+2y=0⇒x−y=0S_1-S_2=0\Rightarrow(3x+5y)-(5x+3y)=0\Rightarrow-2x+2y=0\Rightarrow x-y=0.

Centre of S1S_1 is (−32,−52)\left(-\dfrac32,-\dfrac52\right), and r12=(32)2+(52)2−4=94+254−4=92r_1^2=\left(\dfrac32\right)^2+\left(\dfrac52\right)^2-4=\dfrac94+\dfrac{25}{4}-4=\dfrac92.

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