Question

please give me answer to the question
please ​

Answer 1

SolutioN

Let,

• Centre of circle C ( 2 , 2 )
• Point passes through ( 4 , 5 )

$\tt \longrightarrow {(x - h)}^{2} + {(y - k)}^{2} = {R}^{2}$

Let's find Radius OR distance of ( OA )

$\tt \longrightarrow \sqrt{ {(2 - 4)}^{2} + {(2 - 5)}^{2} }$

$\tt \longrightarrow \sqrt{13} .$

_________________________________

$\tt \longrightarrow {(x - 2)}^{2} + {(y - 2)}^{2} = { \Big(\sqrt{13} \Big)}^{2}$

$\tt \longrightarrow {(x - 2)}^{2} + {(y - 2)}^{2} = 13.$

$\tt \longrightarrow {x}^{2} + {y}^{2} - 4x - 4y - 5 = 0.$

MorE Information About Circle.

Equation of circle.

$\tt \: \: \: \: \: \: \: \: x^2 + {y}^{2} + 2gx + 2fy + c = 0.$

$\tt \: \: \: \: \: \: \: \bullet \: e = 0.$

$\tt \: \: \: \: \: \: \: \bullet \: \triangle \: \cancel{=} \: \: 0.$

$\tt \: \: \: \: \: \: \: \bullet \: h = 0.$

$\tt \: \: \: \: \: \: \: \bullet \: a = b.$

Answer 2

Answer:

SolutioN

Let,

Centre of circle C ( 2 , 2 )

Point passes through ( 4 , 5 )

\tt \longrightarrow {(x - h)}^{2} + {(y - k)}^{2} = {R}^{2}⟶(x−h)

2

+(y−k)

2

=R

2

Let's find Radius OR distance of ( OA )

\tt \longrightarrow \sqrt{ {(2 - 4)}^{2} + {(2 - 5)}^{2} }⟶

(2−4)

2

+(2−5)

2

\tt \longrightarrow \sqrt{13} .⟶

13

.

_________________________________

\tt \longrightarrow {(x - 2)}^{2} + {(y - 2)}^{2} = { \Big(\sqrt{13} \Big)}^{2}⟶(x−2)

2

+(y−2)

2

=(

13

)

2

\tt \longrightarrow {(x - 2)}^{2} + {(y - 2)}^{2} = 13.⟶(x−2)

2

+(y−2)

2

=13.

\tt \longrightarrow {x}^{2} + {y}^{2} - 4x - 4y - 5 = 0.⟶x

2

+y

2

−4x−4y−5=0.

MorE Information About Circle.

Equation of circle.

\tt \: \: \: \: \: \: \: \: x^2 + {y}^{2} + 2gx + 2fy + c = 0.x

2

+y

2

+2gx+2fy+c=0.

\tt \: \: \: \: \: \: \: \bullet \: e = 0.∙e=0.

\tt \: \: \: \: \: \: \: \bullet \: \triangle \: \cancel{=} \: \: 0.∙△

=

0.

\tt \: \: \: \: \: \: \: \bullet \: h = 0.∙h=0.

\tt \: \: \: \: \: \: \: \bullet \: a = b.∙a=b.