Respuesta :
Answer:
[tex]{ \tt{line : x = 2y + 5}} \\ { \tt{circle : {x}^{2} + {y}^{2} = 5}} \\ substitute \: line \: in \: circle \\ { \tt{ {(2y + 5)}^{2} + {y}^{2} = 5 }} \\ { \tt{4 {y}^{2} + 20y + 25 + {y}^{2} = 5}} \\ { \tt{5 {y}^{2} + 20y + 20 = 0 }} \\ { \tt{ {y}^{2} + 4y + 4 = 0 }} \\ y = - 2 \\ x = 2( - 2) + 5 = 1 \\ { \tt{centre = ( - 1, \: 2)}} \\ { \tt{radius = 5 \: units}} \\ hence \: proved[/tex]
