A circle passes through the point $(3, 4)$ and cuts the circle $x^{2} + y^{2} = a^{2}$ orthogonally. The locus of its centre is a straight line. If the distance of the straight line from the origin is $817$ then find the value of $(8150 - a^{2})$ .

Open in App

Solution

Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
Since it passes through $(3, 4)$
$\Rightarrow 6 g + 8 f + c = - 25$
As it cuts the circle $x^{2} + y^{2} - a^{2} = 0$ orthogonally
$2 g \times 0 + 2 f \times 0 = c - a^{2}$ ( $∵ 2 g g^{'} + 2 f f^{'} = c + c^{'}$ )
$\Rightarrow 6 g + 8 f + a^{2} + 25 = 0$
Locus of the centre $(- g, - f)$ is $6 x + 8 y - (a^{2} + 25) = 0$
Distance of the line from the origin is
$817 = ∣ ∣ ∣ - \frac{a^{2} + 25}{\sqrt{36 + 64}} ∣ ∣ ∣$
$\Rightarrow a^{2} + 25 = 8170$
$\Rightarrow a^{2} = 8145$

$\Rightarrow 8150 - a^{2} = 5$

Suggest Corrections

Similar questions

(3, 4)

x^{2} + y^{2} = a^{2}

25

a^{2} =

(1, 2)

x^{2} + y^{2} = 4

2 x + 4 y + 9 = 0

(a, b)

x^{2} + y^{2} = 4

m a x + m a y - (a^{2} + b^{2} + p^{2})

m

(a, b)

x^{2} + y^{2} = k^{2}

2 a x + 2 b y - (a^{2} + b^{2} + k^{2}) = 0

(3, 4)

x^{2} + y^{2} = 9

3 x + 4 y = λ

λ =

Join BYJU'S Learning Program