Find Area of triangle

$\text{Points of intersection between line y=x+1 and } \bigcirc x^2+y^2=25~~ are\\ from ~x^2+(x+1)^2=25~\implies x_{1,2}=\dfrac{-2\pm \sqrt{2^2-4(-25)(2)}}{2*2}\\y_{1,2}=x_{1,2}+1=\dfrac{2\pm \sqrt{2^2-4(-25)(2)}}{2*2}\\\implies~(x_1,y_1)=\left(-\dfrac 1 2 + \dfrac{\sqrt{51}}{2} ,~~~ \dfrac 1 2 + \dfrac{\sqrt{51}}{2} \right)... \\ ~(x_2,y_2)=\left(-\dfrac 1 2 - \dfrac{\sqrt{51}}{2} ,~~~ \dfrac 1 2 - \dfrac{\sqrt{51}}{2} \right) \\\therefore~PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}=\sqrt{(\sqrt{51})^2+(\sqrt{51})^2}=\sqrt{102}\\PQ ~slope~is~1. \\\therefore ~it~ makes~45^o ~with~X-axis,~intersects~axes~at~(-1,0), (0,1)\\\therefore \perp to~PQ~from~O=\dfrac {1}{ \sqrt2} ~\implies ~required~area~=\dfrac 1 2 *base*height\\=\dfrac 1 2 *\sqrt{102}*\dfrac {1}{\sqrt 2 }=~~~~~\color{#D61F06}{\huge \boxed{3.571} }$

Please note that this is a very long and ambiguous solution. If you know the shorter way please say me.

Point P and Q are at 5 units from the origin (0,0).

By distance formula

From the question,

x - y + 1 = 0

x - y = -1 -----------------------------(2)

Multiplying equation (2) by (x + y) we get,

If y = -3

then x = y -1 = -3-1= -4

If y = 4

then x = y-1= 4-1 = 3

Hence,

Three points are: P(-3, -4) Q(4, 3) O(0,0)