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Q.Show that x2+y2−6x−9y+13=0x^2 + y^2 - 6x - 9y + 13 = 0, x2+y2−2x−16yx^2 + y^2 - 2x - 16y [... remainder of this circle's equation and a closing clause, most likely ending '= 0 touch each other', are cut off by the source scan's right-edge crop ...]. Find the point of contact and the equation of common tangent at the point of contact.

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 7mImportance★★★★★
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The source scan's right edge crops the second circle's constant term and closing clause. Taking it as x2+y2−2x−16y=0x^2+y^2-2x-16y=0 (matching the printed fragment and the stated conclusion 'touch each other') gives circles whose centre distance exactly equals ∣r1−r2∣|r_1-r_2| — confirming internal tangency at (5,1)(5,1) with common tangent 4x−7y−13=04x-7y-13=0.

Circle 1: x2+y2−6x−9y+13=0⇒x^2+y^2-6x-9y+13=0 \Rightarrow centre C1=(3,4.5)C_1=(3,4.5), r1=32+4.52−13=16.25=652r_1=\sqrt{3^2+4.5^2-13}=\sqrt{16.25}=\frac{\sqrt{65}}{2}.

Circle 2: x2+y2−2x−16y=0⇒x^2+y^2-2x-16y=0 \Rightarrow centre C2=(1,8)C_2=(1,8), r2=12+82−0=65r_2=\sqrt{1^2+8^2-0}=\sqrt{65}.

Distance between centres:

C1C2=(3−1)2+(4.5−8)2=4+12.25=16.25=652C_1C_2=\sqrt{(3-1)^2+(4.5-8)^2}=\sqrt{4+12.25}=\sqrt{16.25}=\frac{\sqrt{65}}{2}

Since C1C2=r2−r1=65−652=652C_1C_2 = r_2-r_1 = \sqrt{65}-\frac{\sqrt{65}}{2}=\frac{\sqrt{65}}{2}, the circles touch internally.

Common tangent at the point of contact (= radical axis, since the circles meet at exactly one point): subtract the equations,

(−6x−9y+13)−(−2x−16y)=0⇒−4x+7y+13=0⇒4x−7y−13=0(-6x-9y+13)-(-2x-16y)=0 \Rightarrow -4x+7y+13=0 \Rightarrow 4x-7y-13=0

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