Let the point at which the circle passing through (0,0) and (1,0) touches the circle x2+y2=9 is P(h , k) , then |k| is equal to

Question

Let the point at which the circle passing through (0,0) and (1,0) touches the circle x2+y2=9 is P(h , k) , then |k| is equal to (A) √5 (B) 2√2 (C) √6 (D) √7

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-mmpxuoy5pnub.png" /> From the figure, for the small circle, the radius is (3/2) units and the centre is ((1/2) , λ ) Therefore, √((1/2))2 + (λ )2=(3/2)⇒ λ =±√2 Since, OP is the diameter of the circle, ⇒ P=(1 , ± 2 √2)⇒ |k|=2√2

Q. Let the point at which the circle passing through $(0, 0)$ and $(1, 0)$ touches the circle $x^{2} + y^{2} = 9$ is $P (h, k)$ , then $∣ k ∣$ is equal to

$5$

$22$

$6$

$7$

From the figure,
for the small circle, the radius is $\frac{3}{2}$ units and the centre is $(\frac{1}{2}, λ)$
Therefore,
$(\frac{1}{2})^{2} + (λ)^{2} = \frac{3}{2} \Rightarrow λ = \pm 2$
Since, $OP$ is the diameter of the circle,
$\Rightarrow P = (1, \pm 22) \Rightarrow ∣ k ∣ = 22$

Q. Let the point at which the circle passing through (0,0) and (1,0) touches the circle x2+y2=9 is P(h,k) , then ∣k∣ is equal to

5​

22​

6​

7​

From the figure, for the small circle, the radius is 23​ units and the centre is (21​,λ) Therefore, (21​)2+(λ)2​=23​⇒λ=±2​ Since, OP is the diameter of the circle, ⇒P=(1,±22​)⇒∣k∣=22​

Q. Let the point at which the circle passing through $(0, 0)$ and $(1, 0)$ touches the circle $x^{2} + y^{2} = 9$ is $P (h, k)$ , then $∣ k ∣$ is equal to

$5$

$22$

$6$

$7$

From the figure,
for the small circle, the radius is $\frac{3}{2}$ units and the centre is $(\frac{1}{2}, λ)$
Therefore,
$(\frac{1}{2})^{2} + (λ)^{2} = \frac{3}{2} \Rightarrow λ = \pm 2$
Since, $OP$ is the diameter of the circle,
$\Rightarrow P = (1, \pm 22) \Rightarrow ∣ k ∣ = 22$