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Question
The equation of the circle whose radius is 5units and which touches the circle, x2+y2−2x−4y−20=0 at the point (5,5) is
The equation of the circle whose radius is 5units and which touches the circle, x2+y2−2x−4y−20=0 at the point (5,5) is
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Solution
The correct option is B x2+y2−18x−16y+120=0
The given circle is x2+y2−2x−4y−20=0 has its centre
−2g=−2 or g=1
−2f=−4 or f=2Thus,thecentreisatC\left(1,2\right)andradius=r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{{1}^{2}+{2}^{2}-\left(-20\right)}=5unitsWehave{O}_{1}=\left(1,2\right)andLet{O}_{2}=\left(\alpha,\beta\right)bethecentresA\left(5,5\right)isthemid−pointof{O}_{1}{O}_{2}\therefore \dfrac{1+\alpha}{2}=5, \dfrac{2+\beta}{2}=5Onsimplifying,weget\alpha=9,\beta=8Thecircleis{\left(x-9\right)}^{2}+{\left(y-8\right)}^{2}={5}^{2}or{x}^{2}+{y}^{2}-18x-16y+120=0$
The given circle is x2+y2−2x−4y−20=0 has its centre
−2g=−2 or g=1
−2f=−4 or f=2Thus,thecentreisatC\left(1,2\right)andradius=r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{{1}^{2}+{2}^{2}-\left(-20\right)}=5unitsWehave{O}_{1}=\left(1,2\right)andLet{O}_{2}=\left(\alpha,\beta\right)bethecentresA\left(5,5\right)isthemid−pointof{O}_{1}{O}_{2}\therefore \dfrac{1+\alpha}{2}=5, \dfrac{2+\beta}{2}=5Onsimplifying,weget\alpha=9,\beta=8Thecircleis{\left(x-9\right)}^{2}+{\left(y-8\right)}^{2}={5}^{2}or{x}^{2}+{y}^{2}-18x-16y+120=0$
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