Question

find equation of sphere which has centre (1,2,3) and passing through (0,0,0)​

Answer 1

The equation of sphere which has centre (1,2,3) and passing through (0,0,0) is x² + y² + z² - 2x - 4y - 6z = 0

Step-by-step explanation:

Given,

The centre O of the sphere $=(1,2,3)$

The sphere passes through the point A $=(0,0,0)$

Therefore, the raius of the sphere

$r=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$r=\sqrt{(1-0)^2+(2-0)^2+(3-0)^2}$

$\implies r=\sqrt{1+4+9}$

$\implies r=\sqrt{14}$

We know that equation of the sphere with centre at $(x_0,y_0,z_0)$ and radius r is

$(x-x_0)^2+(y-y_0)^2+(z-z)0)^2=r^2$

Therefore, the equation of the sphere is

$(x-1)^2+(y-2)^2+(z-3)^2=(\sqrt{14})^2$

$\implies x^2-2x+1+y^2-4y+4+z^2-6z+9=14$

$\implies x^2+y^2+z^2-2x-4y-6z=0$

This is the required equation of the sphere.

Hope this answer is helpful.

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