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Question 28 of 37

Q.Find the equation of the circle which touches the circle x2+y2−2x−4y−20=0x^2 + y^2 - 2x - 4y - 20 = 0 externally at (5,5)(5, 5) with radius 5.

Telangana TsbieTelangana Board of Intermediate Education 2022Subjective· 7mImportance★★★★★
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The centre of the required circle lies on the ray from the given circle's centre through the point of contact, at a distance equal to the sum of the radii.

Given circle: x2+y2−2x−4y−20=0⇒g1=−1,f1=−2,c1=−20x^2+y^2-2x-4y-20=0 \Rightarrow g_1=-1,f_1=-2,c_1=-20, centre C1=(1,2)C_1=(1,2), r1=1+4+20=5r_1=\sqrt{1+4+20}=5.

Check (5,5)(5,5) lies on this circle: 25+25−10−20−20=025+25-10-20-20=0 ✓.

Since the required circle touches this one externally at (5,5)(5,5) with radius r2=5r_2=5, its centre C2C_2 lies on the line from C1C_1 through (5,5)(5,5), at distance r2r_2 beyond (5,5)(5,5).

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