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Question 16 of 37

Q.Show that the circles : x2+y2−6x−9y+13=0x^2 + y^2 - 6x - 9y + 13 = 0, x2+y2−2x−16y=0x^2 + y^2 - 2x - 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Telangana TsbieTelangana Board of Intermediate Education 2018Subjective· 7mImportance★★★★★
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Two circles touch when the distance between centres equals the sum (external touch) or the difference (internal touch) of their radii; the common tangent there is the radical axis S1−S2=0S_1-S_2=0.

S1≡x2+y2−6x−9y+13=0S_1\equiv x^2+y^2-6x-9y+13=0: centre (3,4.5)(3,4.5), r12=9+20.25−13=16.25=654r_1^2=9+20.25-13=16.25=\dfrac{65}{4}, so r1=652r_1=\dfrac{\sqrt{65}}{2}.

S2≡x2+y2−2x−16y=0S_2\equiv x^2+y^2-2x-16y=0: centre (1,8)(1,8), r22=1+64=65r_2^2=1+64=65, so r2=65r_2=\sqrt{65}.

Distance between centres: d=(3−1)2+(4.5−8)2=4+12.25=16.25=652d=\sqrt{(3-1)^2+(4.5-8)^2}=\sqrt{4+12.25}=\sqrt{16.25}=\dfrac{\sqrt{65}}{2}.

Since r2−r1=65−652=652=dr_2-r_1=\sqrt{65}-\dfrac{\sqrt{65}}{2}=\dfrac{\sqrt{65}}{2}=d, the circles touch internally.

Point of contact (dividing the centres' join externally in ratio r1:r2r_1:r_2): …

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