Question

Given points,

(4,1),(6,5)

equation of circle (x-h)2+(y-k)2=r2

- (4-h)2+(1-k)2=r2 (1)

→ (6-h)2+(5-k)2=r2 (2)

solving the above 2 equations, we get,

h+2k=11 (3)

given, 4h+k=16 (4)

solving the above 2 equations, we get,

h=3,K=4

substituting the above values in (1), we get,

(4-3)2+(1-4)2=r2

r=10

Hence, the equation is,

46 14 1 97%

142

8

x 5.0

(x-3)2+(y-4)2=(10)2

X2+y2-6x-8y+15=0​

Answer 1

$\huge \sf \bf {\boxed{\underline {\pink{\underline {✠Aɴʂᴡᴇʀ࿐ :−}}}}}$

Given points,

(4,1),(6,5)

equation of circle (x-h)2+(y-k)2=r2

- (4-h)2+(1-k)2=r2 (1)

→ (6-h)2+(5-k)2=r2 (2)

solving the above 2 equations, we get,

h+2k=11 (3)

given, 4h+k=16 (4)

solving the above 2 equations, we get,

h=3,K=4

substituting the above values in (1), we get,

(4-3)2+(1-4)2=r2

r=10

Hence, the equation is,

46 14 1 97%

142

8

x 5.0

(x-3)2+(y-4)2=(10)2

X2+y2-6x-8y+15=0

Answer 2

Answer:

\huge \sf \bf {\boxed{\underline {\pink{\underline {✠Aɴʂᴡᴇʀ࿐ :−}}}}}✠Aɴʂᴡᴇʀ࿐:−

Given points,

(4,1),(6,5)

equation of circle (x-h)2+(y-k)2=r2

- (4-h)2+(1-k)2=r2 (1)

→ (6-h)2+(5-k)2=r2 (2)

solving the above 2 equations, we get,

h+2k=11 (3)

given, 4h+k=16 (4)

solving the above 2 equations, we get,

h=3,K=4

substituting the above values in (1), we get,

(4-3)2+(1-4)2=r2

r=10

Hence, the equation is,

46 14 1 97%

142

8

x 5.0

(x-3)2+(y-4)2=(10)2

X2+y2-6x-8y+15=0