Which one is a common point to the hyperbola $x^{2} - y^{2} = 7$ and the circle $x^{2} + y^{2} = 25$ ?

Open in App

Solution

$x^{2} - y^{2} = 7$ (1)
$x^{2} + y^{2} = 25$ (2)
(1) + (2)
$\Rightarrow 2 x^{2} = 32$
$x^{2} = 16$
$x = 4, - 4$
$y = 3, - 3$
So the point are
$(4, 3) (- 4, 3)$
$(4, - 3) (- 4, - 3)$

Suggest Corrections

Similar questions

Q. For hyperbola

\frac{x^{2}}{16} - \frac{y^{2}}{12} = 0

and circle

x^{2} + y^{2} = 25

, find concyclic points.

For hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$ , and circle $x^{2} + y^{2} = 100$ , the ray from one of concyclic points passes through a focus.

For hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{12} = 1$ , and circle $x^{2} + y^{2} = 100$ , Four concyclic points are

For hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$ , and circle $x^{2} + y^{2} = 100$ , con-cyclic points are

Q. Tangent drawn from a point on the circle

x^{2} + y^{2} = 11

to the hyperbola

\frac{x^{2}}{36} - \frac{y^{2}}{25} = 1,

then tangents are at angle

Join BYJU'S Learning Program

Which one is a common point to the hyperbola x2−y2=7 and the circle x2+y2=25 ?

x2−y2=7 __(1)x2+y2=25 __(2)(1) + (2)⇒2x2=32x2=16x=4,−4y=3,−3So the point are(4,3)(−4,3)(4,−3)(−4,−3)

Which one is a common point to the hyperbola $x^{2} - y^{2} = 7$ and the circle $x^{2} + y^{2} = 25$ ?

$x^{2} - y^{2} = 7$ (1)
$x^{2} + y^{2} = 25$ (2)
(1) + (2)
$\Rightarrow 2 x^{2} = 32$
$x^{2} = 16$
$x = 4, - 4$
$y = 3, - 3$
So the point are
$(4, 3) (- 4, 3)$
$(4, - 3) (- 4, - 3)$