# Diameter from of the Equation of a Sphere

## Diameter from of the Equation of a Sphere

Let AB be the diameter of a sphere whose centre is C. let the vectors of the extremities A and B of the diameter be $$\vec{a}$$ and $$\vec{b}$$, respectively. Let P be any point on the spheres. Suppose the position vector of P is $$\vec{r}$$. We know that the angle in a hemisphere is right angle. Thus,

$$\angle APB=\frac{\pi }{2}$$,

$$\overrightarrow{AP}.\overrightarrow{BP}=0$$ … (1)

$$\overrightarrow{AP}=\vec{r}-\vec{a}$$ and $$\overrightarrow{BP}=\vec{r}-\vec{b}$$,

From equation (1)

$$\overrightarrow{AP}.\overrightarrow{BP}=0\Rightarrow \left( \vec{r}-\vec{a} \right)\left( \vec{r}-\vec{b} \right)=0$$,

This is required equation of the sphere.

Cartesian Form:

$$\vec{a}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}$$,

$$\vec{b}={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k}$$ and

$$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$,

$$\vec{r}-\vec{a}=\hat{i}\left( x-{{x}_{1}} \right)+\hat{j}\left( y-{{y}_{1}} \right)+\hat{k}\left( z-{{z}_{1}} \right)$$,

$$\vec{r}-\vec{b}=\hat{i}\left( x-{{x}_{2}} \right)+\hat{j}\left( y-{{y}_{2}} \right)+\hat{k}\left( z-{{z}_{2}} \right)$$,

The equation of the sphere. $$\left( \vec{r}-\vec{a} \right).\left( \vec{r}-\vec{b} \right)=0$$,

$$\left( \hat{i}\left( x-{{x}_{1}} \right)+\hat{j}\left( y-{{y}_{1}} \right)+\hat{k}\left( z-{{z}_{1}} \right) \right).\left( \hat{i}\left( x-{{x}_{2}} \right)+\hat{j}\left( y-{{y}_{2}} \right)+\hat{k}\left( z-{{z}_{2}} \right) \right)=0$$,

$$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)+\left( z-{{z}_{1}} \right)\left( z-{{z}_{2}} \right)=0$$,

This is required equation of the sphere.

Example: Find the equation of the sphere described on the joint of point A and B having position vectors $$2\hat{i}+6\hat{j}-7\hat{k}$$ and $$-2\hat{i}+4\hat{j}-3\hat{k}$$, respectively, as the diameter. Find the center and radius of the sphere.

Solution: If point P with position vector $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$ is any point on the sphere.

Given that $$2\hat{i}+6\hat{j}-7\hat{k}$$, $$-2\hat{i}+4\hat{j}-3\hat{k}$$,

$$\overrightarrow{AP}=\left( x-2 \right)\hat{i}+\left( y-6 \right)\hat{j}+\left( z+7 \right)\hat{k}$$,

$$\overrightarrow{BP}=\left( x+2 \right)\hat{i}+\left( y-4 \right)\hat{j}+\left( z+3 \right)\hat{k}$$,

$$\overrightarrow{AP}.\overrightarrow{BP}=0$$,

$$\left( x-2 \right)\left( x+2 \right)+\left( y-6 \right)\left( y-4 \right)+\left( z+7 \right)\left( z+3 \right)=0$$,

$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-10y+10z+41=0$$,

The center if the sphere is (0, 5, -5) and

Radius is $$r=\sqrt{{{0}^{2}}+{{5}^{2}}+{{\left( -5 \right)}^{2}}-41}$$,

$$r=\sqrt{25+25-41}$$ = 3.