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Q.Show that : 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 2020Subjective· 7mImportance★★★★★
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Two circles touch when the distance between centres equals the sum (external contact) or the difference (internal contact) of their radii; here d=∣r2−r1∣d=|r_2-r_1|, so they touch internally, and the common tangent is the radical axis S1−S2=0S_1-S_2=0.

Circle 1: x2+y2−6x−9y+13=0x^2+y^2-6x-9y+13=0, centre C1=(3,92)C_1=\left(3,\tfrac92\right), r12=9+814−13=654⇒r1=652r_1^2=9+\tfrac{81}{4}-13=\tfrac{65}{4}\Rightarrow r_1=\tfrac{\sqrt{65}}{2}.

Circle 2: x2+y2−2x−16y=0x^2+y^2-2x-16y=0, centre C2=(1,8)C_2=(1,8), r22=1+64=65⇒r2=65r_2^2=1+64=65\Rightarrow r_2=\sqrt{65}.

Distance between centres:

d=(3−1)2+(92−8)2=4+494=654=652d=\sqrt{(3-1)^2+\left(\tfrac92-8\right)^2}=\sqrt{4+\tfrac{49}{4}}=\sqrt{\tfrac{65}{4}}=\dfrac{\sqrt{65}}{2}

Since r2−r1=65−652=652=dr_2-r_1 = \sqrt{65}-\tfrac{\sqrt{65}}{2}=\tfrac{\sqrt{65}}{2}=d, we have d=∣r1−r2∣d=|r_1-r_2| — the circles touch internally (one lies inside the other, touching at one point).

Point of contact: it lies on ray C2→C1C_2\to C_1 extended to distance r2r_2 from C2C_2. Direction C2→C1=(2,−72)C_2\to C_1 = (2,-\tfrac72), of length d=652d=\tfrac{\sqrt{65}}2; unit vector =(465,−765)=\left(\tfrac{4}{\sqrt{65}},-\tfrac{7}{\sqrt{65}}\right).

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