## Dark Calculator |

Converters | Constants | Calculator | Analytical Geometry |

Circle | Vectors | Straight Line | Parabola | Ellipse | Hyperbola |

Coordinate Geometry [Triangle] | Coordinate Geometry [Quadrilateral] |

Operations:
| menu OPERATIONS: ×
In case of inputting square roots, press the square root button and activate it. To insert fractions, insert it converting into decimals. For example, \(\frac{3}{4}\) is to be inputted as \(0.75\). But the output will be in decimals. Again, for working with more irrational numbers like \(2+\sqrt3\), you must input it in decimal form using full-stop \((.)\) as the decimal point. \(x^2+y^2+2gx+2fy+c=0\) The circle has center \((-g, -f)\) and radius \(\sqrt{g^2+f^2-c}\). The standard equation of a straight line is \(ax+by+c=0\). Input the coordinate of the points on which the circle passes: |

Consider that we have two equations of a circle: $$f(x,y)=x^2+y^2+2gx+2fy+c=0$$ Now, whatever be the format of the straight line that intersects this circle, shift it into \(ax+by+d=0\) format. Consider the equation of the straight line as function \(g(x,y)\) According to A. R. Khalifa's rule, we can say the equation of the circle that passes through the intersection points of this circle and the straight line is: $$f(x,y)+k.g(x,y)=0$$ The circle defined in the above equation passes through the point \((x_1,y_1)\) as we can see: $$f(x_1,y_1)+k.g(x_1,y_1)=0$$ $$\Rightarrow k=-\frac{f(x_1,y_1)}{g(x_1,y_1)}$$ Using this value of \(k\), we have the final equation of the circle, which is: $$f(x,y)-\frac{f(x_1,y_1)}{g(x_1,y_1)}.g(x,y)=0$$As a matter of fact, check if the circle is indeed intersected by this straight line. The circle has center \((-g,-f)\) and a radius of \(r=\sqrt{g^2+f^2-c}\). As, we can see, the perpendicular distance of the straight line from the center of the circle must be less than the radius of the circle, which leads to: $$ \left |{{\frac{-ag_1-bf_1+d}{\sqrt{a^2+b^2}}}}\right | \lt \left|\sqrt{g^2+f^2-c}\right| $$ Even if the straight line acts as a tangent to the circle, then the tangent is the common tangent to both the circles. Verify, whether the point lies on the circle or on the straight line. The point must not lie upon any of those.