Find the equation of the circle, the coordinates of the end points of one

Question:

Find the equation of the circle, the coordinates of the end points of one of whose diameters are

$A(p, q)$ and $B(r, s)$

 

Solution:

The equation of a circle passing through the coordinates of the end points of diameters is:

$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$

Substituting, values: $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(\mathrm{p}, \mathrm{q}) \&\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(\mathrm{r}, \mathrm{s})$

We get:

$(x-p)(x-r)+(y-q)(y-s)=0$

$\Rightarrow x^{2}-r x-p x+p r+y^{2}-s y-q y+q s=0$

$\Rightarrow x^{2}+y^{2}-(r+p) x-(s+q) y+(p r+q s)=0$

Ans: $x^{2}+y^{2}-(r+p) x-(s+q) y+(p r+q s)=0$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now