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Question 17 of 37

Q.Find the equation of the radical axis of the circles x2+y2+4x+6y−7=0x^2 + y^2 + 4x + 6y - 7 = 0, 4(x2+y2)+8x+12y−9=04(x^2 + y^2) + 8x + 12y - 9 = 0.

Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 2mImportance★★★★★
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The radical axis of two circles S1=0S_1=0 and S2=0S_2=0 (each with unit coefficient of x2+y2x^2+y^2) is S1−S2=0S_1-S_2=0.

S1:x2+y2+4x+6y−7=0S_1: x^2+y^2+4x+6y-7=0

The second circle is 4(x2+y2)+8x+12y−9=04(x^2+y^2)+8x+12y-9=0; divide by 4 to make the coefficient of x2+y2x^2+y^2 equal to 1:

S2:x2+y2+2x+3y−94=0S_2: x^2+y^2+2x+3y-\frac{9}{4}=0

Radical axis: S1−S2=0S_1-S_2=0

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