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A circle \(S\) of radius 2 units lies in the first quadrant and touches both the coordinate axes. The equation of the circle with centre at \((6,5)\) and touching the circle \(S\) externally is
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Verified Answer
The correct answer is:
\(x^2+y^2-12 x-10 y+52=0\)
From figure
Centre of given circle \(\left(C_1\right)=(2,2)\) radius \(=2\) units
Centre of required circle \(\left(C_2\right)=(6,5)\)
\(\begin{array}{rlr}
C_1 C_2 & =r+2 \quad \text { ( } \because \text { from figure) } \\
\therefore \quad & & \\
\therefore \quad r & =r+2
\end{array}\)
\(\therefore\) Required equation of circle having centre at \((6,5)\) and \(r=3\) is
\(\begin{aligned}
(x-6)^2+(y-5)^2 & =3^2 \\
x^2+y^2-12 x-10 y+52 & =0
\end{aligned}\)
\(\therefore\) Hence, answer is (d).
Centre of given circle \(\left(C_1\right)=(2,2)\) radius \(=2\) units
Centre of required circle \(\left(C_2\right)=(6,5)\)
\(\begin{array}{rlr}
C_1 C_2 & =r+2 \quad \text { ( } \because \text { from figure) } \\
\therefore \quad & & \\
\therefore \quad r & =r+2
\end{array}\)
\(\therefore\) Required equation of circle having centre at \((6,5)\) and \(r=3\) is
\(\begin{aligned}
(x-6)^2+(y-5)^2 & =3^2 \\
x^2+y^2-12 x-10 y+52 & =0
\end{aligned}\)
\(\therefore\) Hence, answer is (d).
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