Question

The possible radius of a circle whose centre is at the origin and which internally touches the circle $x^2+y^2-6x-8y+20=0$,is

• A. $2.5+\sqrt { 5 }$
• B. $3.5+\sqrt { 5 }$
• C. $5+\sqrt { 5 }$
• D. $7.5+\sqrt { 5 }$

Answer 1

Answer: (C)

$5+\sqrt { 5 }Centre\rightarrow (0,0)\\ Distance\quad between\quad centres\quad =\quad ({ r }_{ 1 }-{ r }_{ 2 })\\ Let\quad centre\quad of\quad circle\quad be\quad (h,k)\quad and\quad other\quad circle\quad is\quad (3,4)\\ \Rightarrow \sqrt { (h-3{ ) }^{ 2 }+(k-4{ ) }^{ 2 } } =\quad [r-\sqrt { 9+16-20 } ].\\ \Rightarrow \sqrt { (h-3{ ) }^{ 2 }+(k-4{ ) }^{ 2 } } =\quad [r-\sqrt { 5 } ].\\ \because \quad Centre\quad is\quad origin,\\ \therefore \quad (0,0)\\ h=k=0\\ \Rightarrow 5=[r-\sqrt { 5 } ].\\ \Rightarrow 5=r-\sqrt { 5 } .\\ \Rightarrow r=5+\sqrt { 5 }.$