## Dark Calculator |

Converters | Constants | Calculator | Analytical Geometry |

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

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

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Input the equation of the tangent of the circle: The standard equation of a straight line is \(ax+by+c=0\). Input the point of the tangent where the circle has touched it. Inputting either the abscissa or ordinate will work, we would find the rest. But if you input both, check whether the point lie on the straight line or not. Inputting nothing indicates blank ordinate or abscissa (that the calculator will find out) Input the point upon which the circle passes through. |

Consider that the equation of the circle is \(g(x,y)\) and \(f(x,y)\) as the equation of a straight line. We consider \(g(x,y)\) as a point circle that intersects \(f(x,y)\) at some point say (\(m,n)\). A point circle is basically a point, so it must intersect the line at just a single point. Therefore, since the circle formed from the intersection of a circle and a straight line is given by the A. R. Khalifa rule: $$g(x,y)+k. f(x,y)=0$$ where \(k\) is a constant, we consider that this must be the equation of the circle that we need to find out. Now, there are few things to notice. First, the primary circle we considered i. e. \(g(x,y)\) that it is a point circle, which practically is not. Second, how do we find \(k\)?

Now, the solution is, both the tangent and the circle goes through the point (\(m,n)\), which leads to \(g(m,n)=0\) and \(f(m,n)=0\), that does not help finding \(k\). However, we have another point (\(p,q)\) through which none of the circle \(g(x,y)\) or the straight line \(f(x,y)\) passes through, but the circle \(g(x,y)+k.f(x,y)\) does. Therefore, we can find \(k\) from this equation, which is: $$k=-\frac{g(p,q)}{f(p,q)}$$ Now, it is time to consider \(g(x,y)\). Since, it has a radius of \(0\) and it is centered on \((m,n)\), we have $$g(x,y)=x^2+y^2+2mx+2ny+m^2+n^2=0$$ The circle must be centered on \((m,n)\), since it touches the straight line \(f(x,y)\) on this point. If it is centered anywhere else, it gains a non-zero radius, that is contradictory to the condition that we set up at the beginning. Just as the cases before, the straight line has the form \(f(x,y)=ax+by+c=0\). Therefore, the value of \(k\) finally leads to: $$k=-\frac{p^2+q^2+2mp+2nq+m^2+n^2}{ap+bq+c}$$ Finally the equation of the required circle \(z(x,y)=g(x,y)+k. f(x,y)=0\) is: