Note: If the end points of the diameter is (x_1, y_1) and ( x_2, y_2) , then the equation of the circle is given by: (x-x_1)(x-x_2) + ( y - y_1) ( y - y_2) = 0

\displaystyle \textbf{Question 1: }\text{Find the equation of the circle, the endpoints of whose diameter are }(2,-3)
\displaystyle \text{and }(-2,4).\text{ Also find its centre and radius.}
\displaystyle \text{Answer:}
\displaystyle \text{The endpoints of the diameter are }A(2,-3)\text{ and }B(-2,4).
\displaystyle \text{The equation of a circle with }A(x_1,y_1)\text{ and }B(x_2,y_2)\text{ as endpoints of a diameter is}
\displaystyle (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.
\displaystyle \therefore (x-2)(x+2)+(y+3)(y-4)=0
\displaystyle x^2-4+y^2-y-12=0
\displaystyle \therefore x^2+y^2-y-16=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle 2g=0,\qquad 2f=-1,\qquad c=-16.
\displaystyle \therefore g=0,\qquad f=-\frac{1}{2}.
\displaystyle \text{Centre}=(-g,-f)=\left(0,\frac{1}{2}\right).
\displaystyle \text{Radius}=\sqrt{g^2+f^2-c}
\displaystyle =\sqrt{0^2+\left(-\frac{1}{2}\right)^2-(-16)}
\displaystyle =\sqrt{\frac{1}{4}+16}=\frac{\sqrt{65}}{2}.
\displaystyle \therefore \text{The required circle is }x^2+y^2-y-16=0,
\displaystyle \text{with centre }\left(0,\frac{1}{2}\right)\text{ and radius }\frac{\sqrt{65}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of the circle, the endpoints of whose diameter are the centres}
\displaystyle \text{of the circles }x^2+y^2+6x-14y-1=0\text{ and }x^2+y^2-4x+10y-2=0.
\displaystyle \text{Answer:}
\displaystyle \text{For the circle }x^2+y^2+6x-14y-1=0,
\displaystyle 2g=6,\qquad 2f=-14.
\displaystyle \therefore g=3,\qquad f=-7.
\displaystyle \therefore \text{Its centre is }(-g,-f)=(-3,7).
\displaystyle \text{For the circle }x^2+y^2-4x+10y-2=0,
\displaystyle 2g=-4,\qquad 2f=10.
\displaystyle \therefore g=-2,\qquad f=5.
\displaystyle \therefore \text{Its centre is }(-g,-f)=(2,-5).
\displaystyle \text{Thus, the endpoints of the required diameter are }A(-3,7)\text{ and }B(2,-5).
\displaystyle \therefore [x-(-3)](x-2)+(y-7)[y-(-5)]=0
\displaystyle (x+3)(x-2)+(y-7)(y+5)=0
\displaystyle x^2+x-6+y^2-2y-35=0
\displaystyle \therefore x^2+y^2+x-2y-41=0.
\displaystyle \therefore \text{The equation of the required circle is }x^2+y^2+x-2y-41=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The sides of a square are }x=6,\ x=9,\ y=3\text{ and }y=6.\text{ Find the equation}
\displaystyle \text{of a circle drawn on a diagonal of the square as its diameter.}
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of the square are }A(6,3),\ B(9,3),\ C(9,6)\text{ and }D(6,6).
\displaystyle \text{Taking }AC\text{ as the diameter, the equation of the circle is}
\displaystyle (x-6)(x-9)+(y-3)(y-6)=0.
\displaystyle x^2-15x+54+y^2-9y+18=0
\displaystyle \therefore x^2+y^2-15x-9y+72=0.
\displaystyle \text{Taking }BD\text{ as the diameter, the equation of the circle is}
\displaystyle (x-9)(x-6)+(y-3)(y-6)=0.
\displaystyle x^2-15x+54+y^2-9y+18=0
\displaystyle \therefore x^2+y^2-15x-9y+72=0.
\displaystyle \text{Thus, both diagonals give the same circle.}
\displaystyle \therefore \text{The required equation is }x^2+y^2-15x-9y+72=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equation of the circle circumscribing the rectangle whose sides are}
\displaystyle x-3y=4,\quad 3x+y=22,\quad x-3y=14\text{ and }3x+y=62.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the sides are}
\displaystyle x-3y=4\qquad\ldots\ldots\text{(i)}
\displaystyle 3x+y=22\qquad\ldots\ldots\text{(ii)}
\displaystyle x-3y=14\qquad\ldots\ldots\text{(iii)}
\displaystyle 3x+y=62\qquad\ldots\ldots\text{(iv)}
\displaystyle \text{Solving (i) and (ii), we get }A(7,1).
\displaystyle \text{Solving (ii) and (iii), we get }B(8,-2).
\displaystyle \text{Solving (iii) and (iv), we get }C(20,2).
\displaystyle \text{Solving (iv) and (i), we get }D(19,5).
\displaystyle \text{Since the diagonals of a rectangle are diameters of its circumscribed circle,}
\displaystyle \text{taking }AC\text{ as the diameter,}
\displaystyle (x-7)(x-20)+(y-1)(y-2)=0.
\displaystyle x^2-27x+140+y^2-3y+2=0
\displaystyle \therefore x^2+y^2-27x-3y+142=0.
\displaystyle \text{Taking }BD\text{ as the diameter,}
\displaystyle (x-8)(x-19)+(y+2)(y-5)=0.
\displaystyle x^2-27x+152+y^2-3y-10=0
\displaystyle \therefore x^2+y^2-27x-3y+142=0.
\displaystyle \text{Thus, both diagonals give the same circle.}
\displaystyle \therefore \text{The required equation is }x^2+y^2-27x-3y+142=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of the circle passing through the origin and the points where}
\displaystyle \text{the line }3x+4y=12\text{ meets the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }3x+4y=12.
\displaystyle \text{Putting }y=0,\text{ we get }3x=12\Rightarrow x=4.
\displaystyle \therefore \text{The }x\text{-intercept is }A(4,0).
\displaystyle \text{Putting }x=0,\text{ we get }4y=12\Rightarrow y=3.
\displaystyle \therefore \text{The }y\text{-intercept is }B(0,3).
\displaystyle \text{Since }OA\perp OB,\ \angle AOB=90^\circ.
\displaystyle \therefore AB\text{ is a diameter of the required circle.}
\displaystyle \text{Hence, its equation is}
\displaystyle (x-4)(x-0)+(y-0)(y-3)=0.
\displaystyle x^2-4x+y^2-3y=0
\displaystyle \therefore \text{The required equation is }x^2+y^2-4x-3y=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equation of the circle which passes through the origin and cuts off}
\displaystyle \text{intercepts }a\text{ and }b\text{ respectively from the }x\text{- and }y\text{-axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The circle passes through }O(0,0),\ A(a,0)\text{ and }B(0,b).
\displaystyle \text{Since }OA\perp OB,\ \angle AOB=90^\circ.
\displaystyle \therefore AB\text{ is a diameter of the required circle.}
\displaystyle \text{Hence, its equation is}
\displaystyle (x-a)(x-0)+(y-0)(y-b)=0.
\displaystyle x^2-ax+y^2-by=0
\displaystyle \therefore \text{The required equation is }x^2+y^2-ax-by=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the equation of the circle whose diameter is the line segment joining}
\displaystyle (-4,3)\text{ and }(12,-1).\text{ Also find the intercept made by it on the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The endpoints of the diameter are }A(-4,3)\text{ and }B(12,-1).
\displaystyle \text{The equation of a circle with }A(x_1,y_1)\text{ and }B(x_2,y_2)\text{ as endpoints}
\displaystyle \text{of a diameter is }(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.
\displaystyle \therefore [x-(-4)](x-12)+(y-3)[y-(-1)]=0
\displaystyle (x+4)(x-12)+(y-3)(y+1)=0
\displaystyle x^2-8x-48+y^2-2y-3=0
\displaystyle \therefore x^2+y^2-8x-2y-51=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{To find the points where the circle meets the }y\text{-axis, put }x=0\text{ in (i).}
\displaystyle y^2-2y-51=0
\displaystyle y=\frac{2\pm\sqrt{(-2)^2-4(1)(-51)}}{2}
\displaystyle =\frac{2\pm\sqrt{208}}{2}
\displaystyle =\frac{2\pm4\sqrt{13}}{2}=1\pm2\sqrt{13}.
\displaystyle \therefore \text{The points of intersection with the }y\text{-axis are}
\displaystyle P\left(0,1+2\sqrt{13}\right)\text{ and }Q\left(0,1-2\sqrt{13}\right).
\displaystyle \text{Length of the intercept on the }y\text{-axis}
\displaystyle =\left|\left(1+2\sqrt{13}\right)-\left(1-2\sqrt{13}\right)\right|
\displaystyle =4\sqrt{13}\text{ units}.
\displaystyle \therefore \text{The required circle is }x^2+y^2-8x-2y-51=0,
\displaystyle \text{and the intercept made by it on the }y\text{-axis is }4\sqrt{13}\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The abscissae of the two points }A\text{ and }B\text{ are the roots of the equation}
\displaystyle x^2+2ax-b^2=0,\text{ and their ordinates are the roots of the equation}
\displaystyle y^2+2py-q^2=0.\text{ Find the equation of the circle with }AB\text{ as diameter.}
\displaystyle \text{Also find its radius.}
\displaystyle \text{Answer:}
\displaystyle \text{The roots of }x^2+2ax-b^2=0\text{ are}
\displaystyle x=-a\pm\sqrt{a^2+b^2}.
\displaystyle \text{The roots of }y^2+2py-q^2=0\text{ are}
\displaystyle y=-p\pm\sqrt{p^2+q^2}.
\displaystyle \text{Therefore, we may take}
\displaystyle A\left(-a+\sqrt{a^2+b^2},-p+\sqrt{p^2+q^2}\right)
\displaystyle \text{and }B\left(-a-\sqrt{a^2+b^2},-p-\sqrt{p^2+q^2}\right).
\displaystyle \text{The equation of the circle with }AB\text{ as diameter is}
\displaystyle \left(x+a-\sqrt{a^2+b^2}\right)\left(x+a+\sqrt{a^2+b^2}\right)
\displaystyle +\left(y+p-\sqrt{p^2+q^2}\right)\left(y+p+\sqrt{p^2+q^2}\right)=0.
\displaystyle (x+a)^2-(a^2+b^2)+(y+p)^2-(p^2+q^2)=0
\displaystyle x^2+2ax+a^2-a^2-b^2+y^2+2py+p^2-p^2-q^2=0
\displaystyle \therefore x^2+y^2+2ax+2py-b^2-q^2=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle g=a,\qquad f=p,\qquad c=-b^2-q^2.
\displaystyle \therefore \text{Radius}=\sqrt{g^2+f^2-c}
\displaystyle =\sqrt{a^2+p^2-(-b^2-q^2)}
\displaystyle =\sqrt{a^2+b^2+p^2+q^2}.
\displaystyle \therefore \text{The required circle is }x^2+y^2+2ax+2py-b^2-q^2=0,
\displaystyle \text{and its radius is }\sqrt{a^2+b^2+p^2+q^2}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }ABCD\text{ is a square of side }a,\text{ taking }AB\text{ and }AD\text{ as the }x\text{- and }y\text{-axes}
\displaystyle \text{respectively, prove that the equation of the circle circumscribing the square is}
\displaystyle x^2+y^2-a(x+y)=0.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB\text{ and }AD\text{ are taken as the }x\text{- and }y\text{-axes respectively,}
\displaystyle \text{the coordinates of the vertices are }A(0,0),\ B(a,0),\ C(a,a)\text{ and }D(0,a).
\displaystyle \text{The diagonals of a square are diameters of its circumcircle.}
\displaystyle \text{Taking }BD\text{ as the diameter, the equation of the circle is}
\displaystyle (x-a)(x-0)+(y-0)(y-a)=0.
\displaystyle x(x-a)+y(y-a)=0
\displaystyle x^2-ax+y^2-ay=0
\displaystyle \therefore x^2+y^2-a(x+y)=0.
\displaystyle \text{Alternatively, taking }AC\text{ as the diameter,}
\displaystyle (x-0)(x-a)+(y-0)(y-a)=0
\displaystyle x(x-a)+y(y-a)=0
\displaystyle x^2-ax+y^2-ay=0
\displaystyle \therefore x^2+y^2-a(x+y)=0.
\displaystyle \therefore \text{The equation of the circle circumscribing the square is }x^2+y^2-a(x+y)=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The line }2x-y+6=0\text{ meets the circle }x^2+y^2-2y-9=0
\displaystyle \text{at }A\text{ and }B.\text{ Find the equation of the circle on }AB\text{ as diameter.}
\displaystyle \text{Answer:}
\displaystyle \text{The given equations are}
\displaystyle 2x-y+6=0\qquad\ldots\ldots\text{(i)}
\displaystyle x^2+y^2-2y-9=0\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{From (i), }y=2x+6.
\displaystyle \text{Substituting }y=2x+6\text{ in (ii),}
\displaystyle x^2+(2x+6)^2-2(2x+6)-9=0
\displaystyle x^2+4x^2+24x+36-4x-12-9=0
\displaystyle 5x^2+20x+15=0
\displaystyle x^2+4x+3=0
\displaystyle (x+3)(x+1)=0
\displaystyle \therefore x=-3\text{ or }x=-1.
\displaystyle \text{When }x=-3,\quad y=2(-3)+6=0.
\displaystyle \text{When }x=-1,\quad y=2(-1)+6=4.
\displaystyle \therefore A(-3,0)\text{ and }B(-1,4).
\displaystyle \text{The equation of the circle with }AB\text{ as diameter is}
\displaystyle [x-(-3)][x-(-1)]+(y-0)(y-4)=0.
\displaystyle (x+3)(x+1)+y(y-4)=0
\displaystyle x^2+4x+3+y^2-4y=0
\displaystyle \therefore \text{The required equation is }x^2+y^2+4x-4y+3=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the equation of the circle which circumscribes the triangle formed by}
\displaystyle \text{the lines }x=0,\ y=0\text{ and }lx+my=1.
\displaystyle \text{Answer:}
\displaystyle \text{The given lines are}
\displaystyle x=0\qquad\ldots\ldots\text{(i)}
\displaystyle y=0\qquad\ldots\ldots\text{(ii)}
\displaystyle lx+my=1\qquad\ldots\ldots\text{(iii)}
\displaystyle \text{Solving (i) and (ii), we get }C(0,0).
\displaystyle \text{Solving (i) and (iii), we get }A\left(0,\frac{1}{m}\right).
\displaystyle \text{Solving (ii) and (iii), we get }B\left(\frac{1}{l},0\right).
\displaystyle \text{Since }CA\perp CB,\ \angle ACB=90^\circ.
\displaystyle \therefore AB\text{ is a diameter of the circumcircle.}
\displaystyle \text{Hence, the equation of the circle with }AB\text{ as diameter is}
\displaystyle \left(x-0\right)\left(x-\frac{1}{l}\right)+\left(y-\frac{1}{m}\right)(y-0)=0.
\displaystyle x^2-\frac{x}{l}+y^2-\frac{y}{m}=0
\displaystyle \therefore \text{The required equation is }x^2+y^2-\frac{x}{l}-\frac{y}{m}=0.
\displaystyle \text{Equivalently, }lm(x^2+y^2)-mx-ly=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the equations of the circles which pass through the origin and cut off}
\displaystyle \text{equal chords of }\sqrt{2}\text{ units from the lines }y=x\text{ and }y=-x.
\displaystyle \text{Answer:} 2021-02-19_10-38-31\displaystyle \text{The general equation of a circle passing through the origin is}
\displaystyle x^2+y^2+2gx+2fy=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{To find the chord cut from }y=x,\text{ put }y=x\text{ in (i).}
\displaystyle x^2+x^2+2gx+2fx=0
\displaystyle 2x\left[x+(g+f)\right]=0.
\displaystyle \therefore x=0\text{ or }x=-(g+f).
\displaystyle \text{Thus, the points of intersection are }O(0,0)\text{ and}
\displaystyle P\left(-(g+f),-(g+f)\right).
\displaystyle OP=\sqrt{[-(g+f)]^2+[-(g+f)]^2}
\displaystyle =\sqrt{2}\,|g+f|.
\displaystyle \text{Since the chord has length }\sqrt{2},
\displaystyle \sqrt{2}\,|g+f|=\sqrt{2}
\displaystyle \therefore |g+f|=1.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{To find the chord cut from }y=-x,\text{ put }y=-x\text{ in (i).}
\displaystyle x^2+x^2+2gx-2fx=0
\displaystyle 2x\left[x+(g-f)\right]=0.
\displaystyle \therefore x=0\text{ or }x=-(g-f).
\displaystyle \text{Thus, the other point of intersection is}
\displaystyle Q\left(-(g-f),g-f\right).
\displaystyle OQ=\sqrt{[-(g-f)]^2+(g-f)^2}
\displaystyle =\sqrt{2}\,|g-f|.
\displaystyle \text{Since this chord also has length }\sqrt{2},
\displaystyle \sqrt{2}\,|g-f|=\sqrt{2}
\displaystyle \therefore |g-f|=1.\qquad\ldots\ldots\text{(iii)}
\displaystyle \text{From (ii) and (iii),}
\displaystyle g+f=\pm1\text{ and }g-f=\pm1.
\displaystyle \text{The possible values of }(g,f)\text{ are}
\displaystyle (1,0),\ (-1,0),\ (0,1)\text{ and }(0,-1).
\displaystyle \text{Substituting these values in (i), the required circles are}
\displaystyle x^2+y^2+2x=0,
\displaystyle x^2+y^2-2x=0,
\displaystyle x^2+y^2+2y=0,
\displaystyle x^2+y^2-2y=0.
\displaystyle \therefore \text{The equations of the four required circles are}
\displaystyle x^2+y^2\pm2x=0\text{ and }x^2+y^2\pm2y=0.
\displaystyle \\

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