Note: The general equation of circle is \displaystyle  x^2 + y^2 + 2gx + 2fy + c =0 where the center is \displaystyle  ( -g, -f) and radius \displaystyle  = \sqrt{g^2 +f^2 - c}

\displaystyle \textbf{Question 1: }\text{Find the coordinates of the centre and radius of each of} \\ \text{the following circles:}
\displaystyle \text{i) }x^2+y^2+6x-8y-24=0\qquad\text{ii) }2x^2+2y^2-3x+5y=7
\displaystyle \text{iii) }\frac12(x^2+y^2)+x\cos\theta+y\sin\theta-4=0\qquad\text{iv) }x^2+y^2-ax-by=0
\displaystyle \text{Answer:}

\displaystyle \text{i) Given equation }x^2+y^2+6x-8y-24=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,\text{ we get}
\displaystyle g=3,\qquad f=-4,\qquad c=-24.
\displaystyle \therefore \text{Centre}=(-g,-f)=(-3,4).
\displaystyle \text{Radius}=\sqrt{g^2+f^2-c}=\sqrt{3^2+(-4)^2-(-24)}=\sqrt{49}=7.

\displaystyle \text{ii) Given equation }2x^2+2y^2-3x+5y=7.
\displaystyle \Rightarrow x^2+y^2-\frac32x+\frac52y-\frac72=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,\text{ we get}
\displaystyle g=-\frac34,\qquad f=\frac54,\qquad c=-\frac72.
\displaystyle \therefore \text{Centre}=(-g,-f)=\left(\frac34,-\frac54\right).
\displaystyle \text{Radius}=\sqrt{g^2+f^2-c}=\sqrt{\left(-\frac34\right)^2+\left(\frac54\right)^2-\left(-\frac72\right)}=\sqrt{\frac9{16}+\frac{25}{16}+\frac72}=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}4.

\displaystyle \text{iii) Given equation }\frac12(x^2+y^2)+x\cos\theta+y\sin\theta-4=0.
\displaystyle \Rightarrow x^2+y^2+2x\cos\theta+2y\sin\theta-8=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,\text{ we get}
\displaystyle g=\cos\theta,\qquad f=\sin\theta,\qquad c=-8.
\displaystyle \therefore \text{Centre}=(-g,-f)=(-\cos\theta,-\sin\theta).
\displaystyle \text{Radius}=\sqrt{g^2+f^2-c}=\sqrt{\cos^2\theta+\sin^2\theta+8}=\sqrt{1+8}=3.

\displaystyle \text{iv) Given equation }x^2+y^2-ax-by=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,\text{ we get}
\displaystyle g=-\frac a2,\qquad f=-\frac b2,\qquad c=0.
\displaystyle \therefore \text{Centre}=(-g,-f)=\left(\frac a2,\frac b2\right).
\displaystyle \text{Radius}=\sqrt{g^2+f^2-c}=\sqrt{\left(-\frac a2\right)^2+\left(-\frac b2\right)^2}=\frac{\sqrt{a^2+b^2}}2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of the circle passing through the} \\ \text{following points:}
\displaystyle \text{i) }(5,7),(8,1),(1,3)\qquad\text{ii) }(1,2),(3,-4),(5,-6)
\displaystyle \text{iii) }(5,-8),(-2,9),(2,1)\qquad\text{iv) }(0,0),(-2,1),(-3,2)
\displaystyle \text{Answer:}
\displaystyle \text{i) Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }(5,7),(8,1)\text{ and }(1,3)\text{, we get}
\displaystyle 10g+14f+c=-74,\qquad16g+2f+c=-65,\qquad2g+6f+c=-10.
\displaystyle \text{Solving, }g=-\frac{29}{6},\qquad f=-\frac{19}{6},\qquad c=\frac{56}{3}.
\displaystyle \therefore x^2+y^2-\frac{29}{3}x-\frac{19}{3}y+\frac{56}{3}=0.
\displaystyle \text{or }3x^2+3y^2-29x-19y+56=0.
\displaystyle \text{ii) Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }(1,2),(3,-4)\text{ and }(5,-6)\text{, we get}
\displaystyle 2g+4f+c=-5,\qquad6g-8f+c=-25,\qquad10g-12f+c=-61.
\displaystyle \text{Solving, }g=-11,\qquad f=-2,\qquad c=25.
\displaystyle \therefore x^2+y^2-22x-4y+25=0.
\displaystyle \text{iii) Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }(5,-8),(-2,9)\text{ and }(2,1)\text{, we get}
\displaystyle 10g-16f+c=-89,\qquad-4g+18f+c=-85,\qquad4g+2f+c=-5.
\displaystyle \text{Solving, }g=58,\qquad f=24,\qquad c=-285.
\displaystyle \therefore x^2+y^2+116x+48y-285=0.
\displaystyle \text{iv) Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }(0,0),(-2,1)\text{ and }(-3,2)\text{, we get}
\displaystyle c=0,\qquad-4g+2f=-5,\qquad-6g+4f=-13.
\displaystyle \text{Solving, }g=-\frac32,\qquad f=-\frac{11}{2},\qquad c=0.
\displaystyle \therefore x^2+y^2-3x-11y=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equation of the circle passing through }
\displaystyle (3,-2) \text{ and }(-2,0), \ \text{with its centre on the line }2x-y=3.
\displaystyle \text{Answer:}
\displaystyle \text{The circle passes through }(3,-2)\text{ and }(-2,0),\text{ and its centre lies on }2x-y=3.
\displaystyle \text{Let the equation of the circle be}
\displaystyle x^2+y^2+2gx+2fy+c=0.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Substituting }(3,-2)\text{ in i), we get}
\displaystyle 9+4+6g-4f+c=0
\displaystyle \Rightarrow 6g-4f+c=-13.\qquad\text{... ... ... ... ... ii)}
\displaystyle \text{Substituting }(-2,0)\text{ in i), we get}
\displaystyle 4-4g+c=0
\displaystyle \Rightarrow -4g+c=-4.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{The centre of the circle is }(-g,-f).
\displaystyle \text{Since the centre lies on }2x-y=3,
\displaystyle 2(-g)-(-f)=3
\displaystyle \Rightarrow -2g+f=3.\qquad\text{... ... ... ... ... iv)}
\displaystyle \text{Solving ii), iii) and iv), we get}
\displaystyle g=\frac32,\qquad f=6,\qquad c=2.
\displaystyle \text{Therefore, the equation of the required circle is}
\displaystyle x^2+y^2+3x+12y+2=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equation of the circle passing through }
\displaystyle (3,7) \text{ and }(5,5),  \ \text{with its centre on the line }x-4y=1.
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the required circle be}
\displaystyle x^2+y^2+2gx+2fy+c=0.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Substituting }(3,7)\text{ in i), we get}
\displaystyle 9+49+6g+14f+c=0
\displaystyle \Rightarrow 6g+14f+c=-58.\qquad\text{... ... ... ... ... ii)}
\displaystyle \text{Substituting }(5,5)\text{ in i), we get}
\displaystyle 25+25+10g+10f+c=0
\displaystyle \Rightarrow 10g+10f+c=-50.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{The centre of the circle is }(-g,-f).
\displaystyle \text{Since the centre lies on }x-4y=1,
\displaystyle -g-4(-f)=1
\displaystyle \Rightarrow -g+4f=1.\qquad\text{... ... ... ... ... iv)}
\displaystyle \text{Solving ii), iii) and iv), we get}
\displaystyle g=3,\qquad f=1,\qquad c=-90.
\displaystyle \text{Therefore, the equation of the required circle is}
\displaystyle x^2+y^2+6x+2y-90=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that the points }(3,-2),(1,0),(-1,-2)\text{ and }(1,-4) \\ \text{ are concyclic.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(3,-2),\ Q(1,0),\ R(-1,-2)\text{ and }S(1,-4).
\displaystyle \text{Let the equation of the circle be}
\displaystyle x^2+y^2+2gx+2fy+c=0.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Substituting }P(3,-2)\text{ in i), we get}
\displaystyle 9+4+6g-4f+c=0
\displaystyle \Rightarrow 6g-4f+c=-13.\qquad\text{... ... ... ... ... ii)}
\displaystyle \text{Substituting }Q(1,0)\text{ in i), we get}
\displaystyle 1+2g+c=0
\displaystyle \Rightarrow 2g+c=-1.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{Substituting }R(-1,-2)\text{ in i), we get}
\displaystyle 1+4-2g-4f+c=0
\displaystyle \Rightarrow -2g-4f+c=-5.\qquad\text{... ... ... ... ... iv)}
\displaystyle \text{Solving ii), iii) and iv), we get}
\displaystyle g=-1,\qquad f=-2,\qquad c=1.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle x^2+y^2-2x-4y+1=0.\qquad\text{... ... ... ... ... v)}
\displaystyle \text{Now substitute }S(1,-4)\text{ in v):}
\displaystyle 1+16-2-16+1=0.
\displaystyle \therefore S(1,-4)\text{ lies on the circle.}
\displaystyle \therefore P(3,-2),\ Q(1,0),\ R(-1,-2)\text{ and }S(1,-4)\text{ are concyclic.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Show that the points }(5,5),(6,4),(-2,4)\text{ and }
\displaystyle (7,1) \text{ all lie on a circle,} \ \text{and find its equation, centre and radius.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(5,5),\ Q(6,4),\ R(-2,4)\text{ and }S(7,1).
\displaystyle \text{Let the equation of the circle be}
\displaystyle x^2+y^2+2gx+2fy+c=0.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Substituting }P(5,5)\text{ in i), we get}
\displaystyle 25+25+10g+10f+c=0
\displaystyle \Rightarrow 10g+10f+c=-50.\qquad\text{... ... ... ... ... ii)}
\displaystyle \text{Substituting }Q(6,4)\text{ in i), we get}
\displaystyle 36+16+12g+8f+c=0
\displaystyle \Rightarrow 12g+8f+c=-52.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{Substituting }R(-2,4)\text{ in i), we get}
\displaystyle 4+16-4g+8f+c=0
\displaystyle \Rightarrow -4g+8f+c=-20.\qquad\text{... ... ... ... ... iv)}
\displaystyle \text{Solving ii), iii) and iv), we get}
\displaystyle g=-2,\qquad f=-1,\qquad c=-20.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle x^2+y^2-4x-2y-20=0.\qquad\text{... ... ... ... ... v)}
\displaystyle \text{Now substitute }S(7,1)\text{ in v):}
\displaystyle 7^2+1^2-4(7)-2(1)-20=0.
\displaystyle \therefore S(7,1)\text{ lies on the circle.}
\displaystyle \therefore P(5,5),\ Q(6,4),\ R(-2,4)\text{ and }S(7,1)\text{ are concyclic.}
\displaystyle \text{The centre of the circle is }(-g,-f)=(2,1).
\displaystyle \text{The radius is }\sqrt{g^2+f^2-c}=\sqrt{4+1+20}=\sqrt{25}=5\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the equation of the circle circumscribing the triangle} \\ \text{formed by the lines:}
\displaystyle \text{i) }x+y+3=0,\ x-y+1=0,\ x=3
\displaystyle \text{ii) }2x+y-3=0,\ x+y-1=0,\ 3x+2y-5=0
\displaystyle \text{iii) }x+y=2,\ 3x-4y=6,\ x-y=0
\displaystyle \text{iv) }y=x+2,\ 3y=4x,\ 2y=3x
\displaystyle \text{Answer:}

\displaystyle \text{i) The given lines are}
\displaystyle x+y=-3,\qquad x-y=-1,\qquad x=3.
\displaystyle \text{Their pairwise intersections are }A(-2,-1),\ B(3,4)\text{ and }C(3,-6).
\displaystyle \text{Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }A(-2,-1),\ B(3,4)\text{ and }C(3,-6),\text{ we get}
\displaystyle -4g-2f+c=-5,\qquad6g+8f+c=-25,\qquad6g-12f+c=-45.
\displaystyle \text{Solving, }g=-3,\qquad f=1,\qquad c=-15.
\displaystyle \therefore x^2+y^2-6x+2y-15=0.

\displaystyle \text{ii) The given lines are}
\displaystyle 2x+y=3,\qquad x+y=1,\qquad3x+2y=5.
\displaystyle \text{Their pairwise intersections are }A(2,-1),\ B(3,-2)\text{ and }C(1,1).
\displaystyle \text{Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }A(2,-1),\ B(3,-2)\text{ and }C(1,1),\text{ we get}
\displaystyle 4g-2f+c=-5,\qquad6g-4f+c=-13,\qquad2g+2f+c=-2.
\displaystyle \text{Solving, }g=-\frac{13}{2},\qquad f=-\frac52,\qquad c=16.
\displaystyle \therefore x^2+y^2-13x-5y+16=0.

\displaystyle \text{iii) The given lines are}
\displaystyle x+y=2,\qquad3x-4y=6,\qquad x-y=0.
\displaystyle \text{Their pairwise intersections are }A(2,0),\ B(-6,-6)\text{ and }C(1,1).
\displaystyle \text{Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }A(2,0),\ B(-6,-6)\text{ and }C(1,1),\text{ we get}
\displaystyle 4g+c=-4,\qquad-12g-12f+c=-72,\qquad2g+2f+c=-2.
\displaystyle \text{Solving, }g=2,\qquad f=3,\qquad c=-12.
\displaystyle \therefore x^2+y^2+4x+6y-12=0.

\displaystyle \text{iv) The given lines are}
\displaystyle -x+y=2,\qquad-4x+3y=0,\qquad-3x+2y=0.
\displaystyle \text{Their pairwise intersections are }A(6,8),\ B(4,6)\text{ and }C(0,0).
\displaystyle \text{Let the required circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Substituting }A(6,8),\ B(4,6)\text{ and }C(0,0),\text{ we get}
\displaystyle 12g+16f+c=-100,\qquad8g+12f+c=-52,\qquad c=0.
\displaystyle \text{Solving, }g=-23,\qquad f=11,\qquad c=0.
\displaystyle \therefore x^2+y^2-46x+22y=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove that the centres of the circles }x^2+y^2-4x-6y-12=0,
\displaystyle x^2+y^2+2x+4y-10=0 \ \text{and }x^2+y^2-10x-16y-1=0\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-4x-6y-12=0\qquad\Rightarrow\qquad\text{Centre }=(2,3).
\displaystyle x^2+y^2+2x+4y-10=0\qquad\Rightarrow\qquad\text{Centre }=(-1,-2).
\displaystyle x^2+y^2-10x-16y-1=0\qquad\Rightarrow\qquad\text{Centre }=(5,8).
\displaystyle \text{Area of the triangle formed by the three centres}
\displaystyle =\frac12\left|2(-2-8)+(-1)(8-3)+5(3+2)\right|
\displaystyle =\frac12|-20-5+25|=0.
\displaystyle \therefore\text{The three centres are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that the radii of the circles }x^2+y^2=1, 
\displaystyle x^2+y^2-2x-6y-6=0 \ \text{and }x^2+y^2-4x-12y-9=0\text{ are in A.P.}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2=1\qquad\Rightarrow\qquad r_1=1.
\displaystyle x^2+y^2-2x-6y-6=0.
\displaystyle g=-1,\qquad f=-3,\qquad c=-6.
\displaystyle r_2=\sqrt{g^2+f^2-c}=\sqrt{(-1)^2+(-3)^2-(-6)}=\sqrt{16}=4.
\displaystyle x^2+y^2-4x-12y-9=0.
\displaystyle g=-2,\qquad f=-6,\qquad c=-9.
\displaystyle r_3=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+(-6)^2-(-9)}=\sqrt{49}=7.
\displaystyle 2r_2=2(4)=8=r_1+r_3=1+7.
\displaystyle \therefore r_1,\ r_2,\ r_3\text{ are in A.P.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the equation of the circle passing through the origin}
\displaystyle \text{and cutting off chords of lengths }4\text{ and }6\text{ on the positive sides of the } \\ x\text{-axis and } y\text{-axis, respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the circle meet the positive }x\text{-axis at }A(4,0)\text{ and the positive } \\ y\text{-axis at }B(0,6).
\displaystyle \text{Since }OA\perp OB,\ \angle AOB=90^\circ.
\displaystyle \therefore AB\text{ is a diameter of the circle.}
\displaystyle \text{Hence, the centre is the midpoint of }AB.
\displaystyle C=\left(\frac{4+0}{2},\frac{0+6}{2}\right)=(2,3).
\displaystyle \text{Since the circle passes through the origin, its equation is}
\displaystyle x^2+y^2+2gx+2fy=0.
\displaystyle \text{Here, }(-g,-f)=(2,3),\text{ so }g=-2\text{ and }f=-3.
\displaystyle \therefore x^2+y^2-4x-6y=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the equation of the circle concentric with } 
\displaystyle x^2+y^2-6x+12y+15=0 \ \text{and having twice its area.}
\displaystyle \text{Answer:}
\displaystyle \text{The given circle is }x^2+y^2-6x+12y+15=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,\text{ we get}
\displaystyle g=-3,\qquad f=6,\qquad c=15.
\displaystyle \therefore \text{Centre}=(-g,-f)=(3,-6).
\displaystyle \text{The radius of the given circle is}
\displaystyle r_1=\sqrt{g^2+f^2-c}=\sqrt{(-3)^2+6^2-15}=\sqrt{30}.
\displaystyle \text{Let }r\text{ be the radius of the required circle. Since its area is twice the original area,}
\displaystyle \pi r^2=2\pi r_1^2
\displaystyle \Rightarrow r^2=2(30)=60
\displaystyle \Rightarrow r=2\sqrt{15}.
\displaystyle \text{The centre remains }(3,-6).\text{ Therefore, the required equation is}
\displaystyle (x-3)^2+(y+6)^2=60
\displaystyle \Rightarrow x^2-6x+9+y^2+12y+36=60
\displaystyle \therefore x^2+y^2-6x+12y-15=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the equations of the circles passing through }(1,1) \text{ and }
\displaystyle (2,2)\text{ and having radius }1. \ \text{Show that there are two such circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the centre of the circle be }(h,k).
\displaystyle \text{Since the radius is }1,\text{ its equation is}
\displaystyle (x-h)^2+(y-k)^2=1.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Since the circle passes through }(1,1)\text{ and }(2,2),
\displaystyle (1-h)^2+(1-k)^2=1.\qquad\text{... ... ... ... ... ii)}
\displaystyle (2-h)^2+(2-k)^2=1.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{Equating the left sides of ii) and iii), we get}
\displaystyle (1-h)^2+(1-k)^2=(2-h)^2+(2-k)^2
\displaystyle \Rightarrow 2-2h-2k=8-4h-4k
\displaystyle \Rightarrow 2h+2k=6
\displaystyle \Rightarrow h+k=3
\displaystyle \Rightarrow h=3-k.\qquad\text{... ... ... ... ... iv)}
\displaystyle \text{Substituting iv) in ii), we get}
\displaystyle [1-(3-k)]^2+(1-k)^2=1
\displaystyle \Rightarrow (k-2)^2+(1-k)^2=1
\displaystyle \Rightarrow k^2-4k+4+k^2-2k+1=1
\displaystyle \Rightarrow 2k^2-6k+4=0
\displaystyle \Rightarrow k^2-3k+2=0
\displaystyle \Rightarrow (k-1)(k-2)=0
\displaystyle \Rightarrow k=1\text{ or }k=2.
\displaystyle \text{When }k=1,\quad h=3-1=2.
\displaystyle \text{Hence, one centre is }(2,1),\text{ and the corresponding circle is}
\displaystyle (x-2)^2+(y-1)^2=1
\displaystyle \Rightarrow x^2+y^2-4x-2y+4=0.
\displaystyle \text{When }k=2,\quad h=3-2=1.
\displaystyle \text{Hence, the other centre is }(1,2),\text{ and the corresponding circle is}
\displaystyle (x-1)^2+(y-2)^2=1
\displaystyle \Rightarrow x^2+y^2-2x-4y+4=0.
\displaystyle \therefore\text{There are two such circles.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the equation of the circle concentric with } \\ x^2+y^2-4x-6y-3=0\text{ and touching the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The given circle is }x^2+y^2-4x-6y-3=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,\text{ we get}
\displaystyle g=-2,\qquad f=-3.
\displaystyle \therefore\text{Centre}=(-g,-f)=(2,3).
\displaystyle \text{The required circle is concentric with the given circle.}
\displaystyle \therefore\text{Its centre is also }(2,3).
\displaystyle \text{Since it touches the }y\text{-axis, its radius equals the perpendicular distance of the} \\ \text{centre from the }y\text{-axis.}
\displaystyle \therefore r=2.
\displaystyle \text{Hence the required equation is}
\displaystyle (x-2)^2+(y-3)^2=2^2
\displaystyle \Rightarrow x^2+y^2-4x+4+y^2?
\displaystyle \Rightarrow x^2+y^2-4x-6y+9=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If a circle passes through the points }(0,0),\ (a,0)\text{ and }(0,b), \\ \text{ find the coordinates of its centre.}
\displaystyle \text{Answer:}
\displaystyle \text{The required circle passes through }(0,0),\ (a,0)\text{ and }(0,b).
\displaystyle \text{Let its equation be}
\displaystyle x^2+y^2+2gx+2fy+c=0.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Substituting }(0,0)\text{ in i), we get}
\displaystyle c=0.\qquad\text{... ... ... ... ... ii)}
\displaystyle \text{Substituting }(a,0)\text{ in i), we get}
\displaystyle a^2+2ag=0
\displaystyle \Rightarrow a(a+2g)=0
\displaystyle \Rightarrow g=-\frac{a}{2}.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{Substituting }(0,b)\text{ in i), we get}
\displaystyle b^2+2bf=0
\displaystyle \Rightarrow b(b+2f)=0
\displaystyle \Rightarrow f=-\frac{b}{2}.\qquad\text{... ... ... ... ... iv)}
\displaystyle \therefore\text{The centre of the circle is}
\displaystyle (-g,-f)=\left(\frac{a}{2},\frac{b}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the equation of the circle passing through }(2,3)\text{ and } \\ (4,5)\text{ whose centre lies on }y-4x+3=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the circle be}
\displaystyle x^2+y^2+2gx+2fy+c=0.\qquad\text{... ... ... ... ... i)}
\displaystyle \text{Substituting }(2,3)\text{ in i), we get}
\displaystyle 4+9+4g+6f+c=0
\displaystyle \Rightarrow 4g+6f+c=-13.\qquad\text{... ... ... ... ... ii)}
\displaystyle \text{Substituting }(4,5)\text{ in i), we get}
\displaystyle 16+25+8g+10f+c=0
\displaystyle \Rightarrow 8g+10f+c=-41.\qquad\text{... ... ... ... ... iii)}
\displaystyle \text{The centre of the circle is }(-g,-f).
\displaystyle \text{Since the centre lies on }y-4x+3=0,
\displaystyle -f-4(-g)+3=0
\displaystyle \Rightarrow 4g-f=-3.\qquad\text{... ... ... ... ... iv)}
\displaystyle \text{Subtracting ii) from iii), we get}
\displaystyle 4g+4f=-28
\displaystyle \Rightarrow g+f=-7.\qquad\text{... ... ... ... ... v)}
\displaystyle \text{Solving iv) and v), we get}
\displaystyle g=-2,\qquad f=-5.
\displaystyle \text{Substituting these values in ii), we get}
\displaystyle 4(-2)+6(-5)+c=-13
\displaystyle \Rightarrow -8-30+c=-13
\displaystyle \Rightarrow c=25.
\displaystyle \therefore\text{The equation of the required circle is}
\displaystyle x^2+y^2-4x-10y+25=0.
\displaystyle \\

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