The equation of circle of radius √17 unit, with centre on the positive side of x -axis and through the point (0,1) is

Correct answer: (A) x^2+y^2-8 x-1=0. Hint: Let centre be ≡(a, 0)(a>0) So, circle : (x-a)^2+y^2=17 ; as it passes through (0,1) , so a^2+1=17 ⇒ a=4(a ≠-4, a>0) ∴ Equation is : x^2+y^2-8 x-1=0

Q78 / 155 · WBJEE 2020 · Mathematics

The equation of circle of radius $\sqrt{17}$ unit, with centre on the positive side of $x$ -axis and through the point $(0,1)$ is

Correct answer
A
$x^{2}+y^{2}-8 x-1=0$
B
$x^{2}+y^{2}+8 x-1=0$
C
$x^{2}+y^{2}-9 y+1=0$
D
$2 x^{2}+2 y^{2}-3 x+2 y=4$

Solution

Hint: Let centre be $\equiv(a, 0)(a>0)$ So, circle : $(x-a)^{2}+y^{2}=17 ;$ as it passes through $(0,1)$, so $a^{2}+1=17 \quad \Rightarrow a=4(a \neq-4, a>0)$ $\therefore$ Equation is : $x^{2}+y^{2}-8 x-1=0$

Hence, the correct answer is option: 'A'.

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