The area of the circle x2 - 2x + y2 - 10 y + k = 0 is 25 π . The value of k is equal to

Question

The area of the circle x2 - 2x + y2 - 10 y + k = 0 is 25 π . The value of k is equal to (A) -1 (B) 1 (C) 0 (D) 2

Tardigrade Admin · Accepted Answer

We have, x2-2 x+y2-10 y+k=0 ∴ Radius =√g2+f2-c =√(1)2+(5)2-k =√1+25-k =√26-k ∴ Area of circle =π (Radius) 2 ∴ 25 π=π(√26-k)2 ⇒ 25 π =π(26-k) ⇒ 25 =26-k ⇒ k =1

Q. The area of the circle $x^{2} - 2 x + y^{2} - 10 y + k = 0$ is $25 π$ . The value of k is equal to

-1

1

0

2

3

We have,
$x^{2} - 2 x + y^{2} - 10 y + k = 0$
$∴$ Radius $= g^{2} + f^{2} - c$
$= (1)^{2} + (5)^{2} - k$
$= 1 + 25 - k$
$= 26 - k$
$∴$ Area of circle $= π$ (Radius) $^{2}$
$∴ 25 π = π (26 - k)^{2}$
$\Rightarrow 25 π = π (26 - k)$
$\Rightarrow 25 = 26 - k$
$\Rightarrow k = 1$

Q. The area of the circle x2−2x+y2−10y+k=0 is 25π . The value of k is equal to

-1

1

0

2

3

We have, x2−2x+y2−10y+k=0 ∴ Radius =g2+f2−c​ =(1)2+(5)2−k​ =1+25−k​ =26−k​ ∴ Area of circle =π (Radius) 2 ∴25π=π(26−k​)2 ⇒25π=π(26−k) ⇒25=26−k ⇒k=1

Q. The area of the circle $x^{2} - 2 x + y^{2} - 10 y + k = 0$ is $25 π$ . The value of k is equal to

We have,
$x^{2} - 2 x + y^{2} - 10 y + k = 0$
$∴$ Radius $= g^{2} + f^{2} - c$
$= (1)^{2} + (5)^{2} - k$
$= 1 + 25 - k$
$= 26 - k$
$∴$ Area of circle $= π$ (Radius) $^{2}$
$∴ 25 π = π (26 - k)^{2}$
$\Rightarrow 25 π = π (26 - k)$
$\Rightarrow 25 = 26 - k$
$\Rightarrow k = 1$