Question

Find the diameter of circle whose center is (3,2) and passes through (-5,6)

Answer 1

❏Question :-

$\rm Find \: the \: diameter \: of \: circle \: whose \: center \\ \rm is \: (3,2) \: and \: passes \: through \: (-5,6) .$

❏Answer :-

$\to \boxed{ \rm diameter = 8 \sqrt{5} } \:$

❏To Find :-

→ Diameter of circle .

❏Solution :-

According to the question,

Circle has centre c( 3,2) and passes through the point (-5,6) .

We know that ,

→ Equation of circle which passes through centre (h,k) and has radius is r cm .

$\mapsto \boxed{ \rm {(x - h)}^{2} + {(y - k)}^{2} = {r}} \\ \\ \sf we \: have \: \\ \mapsto \rm \: \:( 3,2) \to \: h = 3 \: , \: k = 2 \\ \\ \because \rm \blue{ it \: passes \: through \: the} \: point \: \: ( - 5,6) \\ \: \\ \rm therefore \: , \: r \: is \: the \pink{ \: distance \: between \: } \\ \rm \green{centre \: and \: passing }\: point \: of \: the \: circle \\ \\ \sf hence \\ \rm \red{ r \: is \: the \: distance} \: between \: ( 3,2) \: and \: ( - 5,6) \\ \\ \because \:$

Now ,

$\mapsto \rm \: r = \sqrt{ {( - 5 - 3)}^{2} + {(6 - 2)}^{2} } \\ \\ \mapsto \rm \: r = \sqrt{ {( - 8)}^{2} + 16 } \\ \\ \mapsto \rm \: r = \sqrt{64 + 16} \\ \\ \mapsto \rm \: r = \sqrt{80} \\ \\ \mapsto \: \boxed{\rm \: r = 4\sqrt{5} } \\ \\ \sf \: we \: know \: that \\ \\ \to \rm \: diameter = 2 \times radius \\ \\ \to \rm \: diameter \: = 2 \times 4 \sqrt{5} \\ \\ \to \boxed{ \rm diameter = 8 \sqrt{5} }$