Note: The general equation of a circle is ( x-a)^2 + ( y-b)^2 = r^2 where ( a, b) is the center of the circle and r is the radius of the circle.

\displaystyle \textbf{Question 1: }\text{Find the equation of the circle with:}
\displaystyle \text{(i) Centre }(-2,3)\text{ and radius }4\qquad\text{(ii) Centre }(a,b)\text{ and radius }\sqrt{a^2+b^2}
\displaystyle \text{(iii) Centre }(0,-1)\text{ and radius }1\qquad\text{(iv) Centre }(a\cos\alpha,a\sin\alpha)\text{ and radius }a
\displaystyle \text{(v) Centre }(a,a)\text{ and radius }\sqrt{2}\,a.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a circle with centre }(h,k)\text{ and radius }r\text{ is}
\displaystyle (x-h)^2+(y-k)^2=r^2.

\displaystyle \text{(i) Centre }=(-2,3)\text{ and radius }=4.
\displaystyle [x-(-2)]^2+(y-3)^2=4^2
\displaystyle (x+2)^2+(y-3)^2=16
\displaystyle x^2+4x+4+y^2-6y+9=16
\displaystyle \therefore x^2+y^2+4x-6y-3=0.

\displaystyle \text{(ii) Centre }=(a,b)\text{ and radius }=\sqrt{a^2+b^2}.
\displaystyle (x-a)^2+(y-b)^2=\left(\sqrt{a^2+b^2}\right)^2
\displaystyle x^2-2ax+a^2+y^2-2by+b^2=a^2+b^2
\displaystyle \therefore x^2+y^2-2ax-2by=0.

\displaystyle \text{(iii) Centre }=(0,-1)\text{ and radius }=1.
\displaystyle (x-0)^2+[y-(-1)]^2=1^2
\displaystyle x^2+(y+1)^2=1
\displaystyle x^2+y^2+2y+1=1
\displaystyle \therefore x^2+y^2+2y=0.

\displaystyle \text{(iv) Centre }=(a\cos\alpha,a\sin\alpha)\text{ and radius }=a.
\displaystyle (x-a\cos\alpha)^2+(y-a\sin\alpha)^2=a^2
\displaystyle x^2-2ax\cos\alpha+a^2\cos^2\alpha+y^2-2ay\sin\alpha+a^2\sin^2\alpha=a^2
\displaystyle x^2+y^2-2ax\cos\alpha-2ay\sin\alpha+a^2(\cos^2\alpha+\sin^2\alpha)=a^2
\displaystyle \therefore x^2+y^2-2ax\cos\alpha-2ay\sin\alpha=0.

\displaystyle \text{(v) Centre }=(a,a)\text{ and radius }=\sqrt{2}\,a.
\displaystyle (x-a)^2+(y-a)^2=(\sqrt{2}\,a)^2
\displaystyle x^2-2ax+a^2+y^2-2ay+a^2=2a^2
\displaystyle \therefore x^2+y^2-2ax-2ay=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the centre and radius of each of the following circles:}
\displaystyle \text{(i) }(x-1)^2+y^2=4\qquad\text{(ii) }(x+5)^2+(y+1)^2=9
\displaystyle \text{(iii) }x^2+y^2-4x+6y=5\qquad\text{(iv) }x^2+y^2-x+2y-3=0
\displaystyle \text{Answer:}
\displaystyle \text{The standard equation of a circle is }(x-h)^2+(y-k)^2=r^2,
\displaystyle \text{where }(h,k)\text{ is the centre and }r\text{ is the radius.}

\displaystyle \text{(i) Given }(x-1)^2+y^2=4.
\displaystyle \text{Comparing with }(x-h)^2+(y-k)^2=r^2,
\displaystyle \therefore \text{Centre }=(1,0)\text{ and Radius }=2.

\displaystyle \text{(ii) Given }(x+5)^2+(y+1)^2=9.
\displaystyle \text{Comparing with }(x-h)^2+(y-k)^2=r^2,
\displaystyle \therefore \text{Centre }=(-5,-1)\text{ and Radius }=3.

\displaystyle \text{(iii) Given }x^2+y^2-4x+6y=5.
\displaystyle \Rightarrow (x^2-4x)+(y^2+6y)=5
\displaystyle \Rightarrow (x-2)^2-4+(y+3)^2-9=5
\displaystyle \Rightarrow (x-2)^2+(y+3)^2=18
\displaystyle \text{Comparing with }(x-h)^2+(y-k)^2=r^2,
\displaystyle \therefore \text{Centre }=(2,-3)\text{ and Radius }=\sqrt{18}=3\sqrt{2}.

\displaystyle \text{(iv) Given }x^2+y^2-x+2y-3=0.
\displaystyle \Rightarrow \left(x-\frac12\right)^2-\frac14+(y+1)^2-1-3=0
\displaystyle \Rightarrow \left(x-\frac12\right)^2+(y+1)^2=\frac{17}{4}=\left(\frac{\sqrt{17}}{2}\right)^2
\displaystyle \text{Comparing with }(x-h)^2+(y-k)^2=r^2,
\displaystyle \therefore \text{Centre }=\left(\frac12,-1\right)\text{ and Radius }=\frac{\sqrt{17}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equation of the circle whose centre is }(1,2)\text{ and which passes}
\displaystyle \text{through the point }(4,6).
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the circle is }(1,2)\text{ and it passes through }(4,6).
\displaystyle \text{Therefore, the radius is}
\displaystyle r=\sqrt{(4-1)^2+(6-2)^2}
\displaystyle =\sqrt{9+16}=5.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-1)^2+(y-2)^2=5^2
\displaystyle \Rightarrow x^2-2x+1+y^2-4y+4=25
\displaystyle \therefore x^2+y^2-2x-4y-20=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equation of the circle passing through the point of intersection of}
\displaystyle \text{the lines }x+3y=0\text{ and }2x-7y=0\text{ and whose centre is the point of}
\displaystyle \text{intersection of the lines }x+y+1=0\text{ and }x-2y+4=0.
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the circle is the point of intersection of the lines}
\displaystyle x+y+1=0\qquad\text{and}\qquad x-2y+4=0.
\displaystyle \text{Solving these equations, we get the centre }C=(-2,1).
\displaystyle \text{The circle passes through the point of intersection of the lines}
\displaystyle x+3y=0\qquad\text{and}\qquad 2x-7y=0.
\displaystyle \text{Solving these equations, we get }P=(0,0).
\displaystyle \text{Therefore, the radius is}
\displaystyle r=CP=\sqrt{(-2-0)^2+(1-0)^2}=\sqrt{5}.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle [x-(-2)]^2+(y-1)^2=(\sqrt{5})^2
\displaystyle \Rightarrow (x+2)^2+(y-1)^2=5
\displaystyle \Rightarrow x^2+4x+4+y^2-2y+1=5
\displaystyle \therefore x^2+y^2+4x-2y=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of the circle whose centre lies on the positive direction of}
\displaystyle \text{the }y\text{-axis at a distance }6\text{ from the origin and whose radius is }4.
\displaystyle \text{Answer:}
\displaystyle \text{The centre lies on the positive }y\text{-axis at a distance }6\text{ from the origin.}
\displaystyle \therefore \text{Centre }=(0,6)\text{ and Radius }=4.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-0)^2+(y-6)^2=4^2
\displaystyle \Rightarrow x^2+y^2-12y+36=16
\displaystyle \therefore x^2+y^2-12y+20=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If the equations of two diameters of a circle are }2x+y=6\text{ and}
\displaystyle 3x+2y=4\text{ and the radius is }10,\text{ find the equation of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{The centre of a circle is the point of intersection of any two diameters.}
\displaystyle \text{The equations of the diameters are}
\displaystyle 2x+y=6\qquad\text{and}\qquad 3x+2y=4.
\displaystyle \text{From }2x+y=6,\quad y=6-2x.
\displaystyle \text{Substituting in }3x+2y=4,
\displaystyle 3x+2(6-2x)=4
\displaystyle \Rightarrow 3x+12-4x=4
\displaystyle \Rightarrow x=8.
\displaystyle \therefore y=6-2(8)=-10.
\displaystyle \therefore \text{Centre }=(8,-10)\text{ and Radius }=10.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-8)^2+[y-(-10)]^2=10^2
\displaystyle \Rightarrow (x-8)^2+(y+10)^2=100
\displaystyle \Rightarrow x^2-16x+64+y^2+20y+100=100
\displaystyle \therefore x^2+y^2-16x+20y+64=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the equation of a circle}
\displaystyle \text{(i) which touches both the axes at a distance of }6\text{ units from the origin.}
\displaystyle \text{(ii) which touches the }x\text{-axis at a distance }5\text{ from the origin and has radius }6.
\displaystyle \text{(iii) which touches both the axes and passes through the point }(2,1).
\displaystyle \text{(iv) passing through the origin, having radius }17\text{ and ordinate of the centre }-15.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a circle with centre }(h,k)\text{ and radius }r\text{ is}
\displaystyle (x-h)^2+(y-k)^2=r^2.
\displaystyle \text{(i) The circle touches both axes at points situated }6\text{ units from the origin.}
\displaystyle \text{Therefore, its radius is }6\text{ and its possible centres are}
\displaystyle (6,6),\quad(6,-6),\quad(-6,6),\quad(-6,-6).
\displaystyle \text{Hence the required equations are}
\displaystyle (x-6)^2+(y-6)^2=36
\displaystyle \Rightarrow x^2+y^2-12x-12y+36=0,
\displaystyle (x-6)^2+(y+6)^2=36
\displaystyle \Rightarrow x^2+y^2-12x+12y+36=0,
\displaystyle (x+6)^2+(y-6)^2=36
\displaystyle \Rightarrow x^2+y^2+12x-12y+36=0,
\displaystyle \text{or}\quad (x+6)^2+(y+6)^2=36
\displaystyle \Rightarrow x^2+y^2+12x+12y+36=0.

\displaystyle \text{(ii) The point of contact on the }x\text{-axis is }(5,0)\text{ or }(-5,0).
\displaystyle \text{Since the radius is }6,\text{ the possible centres are}
\displaystyle (5,6),\quad(5,-6),\quad(-5,6),\quad(-5,-6).
\displaystyle \text{Hence the required equations are}
\displaystyle (x-5)^2+(y-6)^2=36
\displaystyle \Rightarrow x^2+y^2-10x-12y+25=0,
\displaystyle (x-5)^2+(y+6)^2=36
\displaystyle \Rightarrow x^2+y^2-10x+12y+25=0,
\displaystyle (x+5)^2+(y-6)^2=36
\displaystyle \Rightarrow x^2+y^2+10x-12y+25=0,
\displaystyle \text{or}\quad (x+5)^2+(y+6)^2=36
\displaystyle \Rightarrow x^2+y^2+10x+12y+25=0.

\displaystyle \text{(iii) Let the radius be }a.\text{ For a circle touching both axes in the first quadrant,}
\displaystyle \text{the centre is }(a,a)\text{ and its equation is}
\displaystyle (x-a)^2+(y-a)^2=a^2.
\displaystyle \Rightarrow x^2+y^2-2ax-2ay+a^2=0.
\displaystyle \text{Since the circle passes through }(2,1),
\displaystyle 2^2+1^2-2a(2)-2a(1)+a^2=0
\displaystyle \Rightarrow a^2-6a+5=0
\displaystyle \Rightarrow (a-1)(a-5)=0
\displaystyle \Rightarrow a=1\text{ or }a=5.
\displaystyle \text{Hence the required equations are}
\displaystyle x^2+y^2-2x-2y+1=0
\displaystyle \text{or}\quad x^2+y^2-10x-10y+25=0.

\displaystyle \text{(iv) Let the centre be }(h,-15)\text{ and the radius be }17.
\displaystyle \text{Since the circle passes through the origin,}
\displaystyle (0-h)^2+[0-(-15)]^2=17^2
\displaystyle \Rightarrow h^2+225=289
\displaystyle \Rightarrow h^2=64
\displaystyle \Rightarrow h=\pm8.
\displaystyle \text{Therefore, the possible centres are }(8,-15)\text{ and }(-8,-15).
\displaystyle \text{Hence the required equations are}
\displaystyle (x-8)^2+(y+15)^2=17^2
\displaystyle \Rightarrow x^2+y^2-16x+30y=0,
\displaystyle \text{or}\quad (x+8)^2+(y+15)^2=17^2
\displaystyle \Rightarrow x^2+y^2+16x+30y=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equation of the circle whose centre is }(3,4)\text{ and which touches}
\displaystyle \text{the straight line }5x+12y-1=0.
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the circle is }(3,4).
\displaystyle \text{The radius is the perpendicular distance of the centre from the given line.}
\displaystyle r=\frac{|5(3)+12(4)-1|}{\sqrt{5^2+12^2}}=\frac{62}{13}.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-3)^2+(y-4)^2=\left(\frac{62}{13}\right)^2
\displaystyle \Rightarrow 169\left(x^2-6x+9+y^2-8y+16\right)=3844
\displaystyle \therefore 169x^2+169y^2-1014x-1352y+381=0.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equation of the circle which touches both the axes and whose centre}
\displaystyle \text{lies on }x-2y=3.
\displaystyle \text{Answer:}
\displaystyle \text{If a circle touches both the axes, its centre is }(\pm r,\pm r),
\displaystyle \text{where }r\text{ is the radius.}
\displaystyle \text{Let the centre be }(\varepsilon_1r,\varepsilon_2r),
\displaystyle \text{where }\varepsilon_1,\varepsilon_2=\pm1.
\displaystyle \text{Since the centre lies on }x-2y=3,
\displaystyle \varepsilon_1r-2\varepsilon_2r=3.
\displaystyle \text{Checking the four sign combinations, the only positive value of }r\text{ is obtained for}
\displaystyle (\varepsilon_1,\varepsilon_2)=(-1,-1).
\displaystyle \Rightarrow -r+2r=3
\displaystyle \Rightarrow r=3.
\displaystyle \therefore \text{Centre }=(-3,-3)\text{ and Radius }=3.
\displaystyle \text{Hence the equation of the circle is}
\displaystyle (x+3)^2+(y+3)^2=3^2
\displaystyle \Rightarrow x^2+6x+9+y^2+6y+9=9
\displaystyle \therefore x^2+y^2+6x+6y+9=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A circle whose centre is the point of intersection of the lines}
\displaystyle 2x-3y+4=0\text{ and }3x+4y-5=0\text{ passes through the origin. Find its equation.}
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the circle is the point of intersection of the lines}
\displaystyle 2x-3y+4=0\qquad\text{and}\qquad 3x+4y-5=0.
\displaystyle \text{Solving these equations, we get}
\displaystyle C=\left(-\frac{1}{17},\frac{22}{17}\right).
\displaystyle \text{Since the circle passes through the origin, its radius is}
\displaystyle r=\sqrt{\left(-\frac{1}{17}\right)^2+\left(\frac{22}{17}\right)^2}
\displaystyle =\sqrt{\frac{1+484}{289}}=\frac{\sqrt{485}}{17}.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle \left(x+\frac{1}{17}\right)^2+\left(y-\frac{22}{17}\right)^2=\frac{485}{289}
\displaystyle \Rightarrow x^2+\frac{2x}{17}+\frac{1}{289}+y^2-\frac{44y}{17}+\frac{484}{289}=\frac{485}{289}
\displaystyle \Rightarrow x^2+y^2+\frac{2x}{17}-\frac{44y}{17}=0
\displaystyle \therefore 17x^2+17y^2+2x-44y=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A circle of radius }4\text{ units touches the coordinate axes in the first quadrant.}
\displaystyle \text{Find the equations of its images with respect to the line mirrors }x=0\text{ and }y=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given circle has radius }4\text{ and touches both axes in the first quadrant.}
\displaystyle \therefore \text{Its centre is }(4,4).
\displaystyle \text{The image of }(4,4)\text{ in the line }y=0\text{ is }(4,-4).
\displaystyle \text{The radius remains unchanged under reflection.}
\displaystyle \therefore \text{The equation of the image circle is}
\displaystyle (x-4)^2+(y+4)^2=4^2
\displaystyle \Rightarrow x^2+y^2-8x+8y+16=0.
\displaystyle \text{The image of }(4,4)\text{ in the line }x=0\text{ is }(-4,4).
\displaystyle \text{The radius remains unchanged under reflection.}
\displaystyle \therefore \text{The equation of the image circle is}
\displaystyle (x+4)^2+(y-4)^2=4^2
\displaystyle \Rightarrow x^2+y^2+8x-8y+16=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the equations of the circles touching the }y\text{-axis at }(0,3)\text{ and}
\displaystyle \text{making an intercept of }8\text{ units on the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the centre of the circle be }C(h,3)\text{ and its radius be }r.
\displaystyle \text{Since the circle touches the }y\text{-axis at }(0,3),
\displaystyle r=|h|.
\displaystyle \text{The circle makes an intercept }AB=8\text{ units on the }x\text{-axis.}
\displaystyle \text{Let }M\text{ be the midpoint of }AB.
\displaystyle \therefore AM=\frac{AB}{2}=4\text{ and }CM=3.
\displaystyle \text{In right-angled }\triangle CMA,
\displaystyle CA^2=CM^2+AM^2
\displaystyle \Rightarrow r^2=3^2+4^2=25
\displaystyle \Rightarrow r=5.
\displaystyle \text{Therefore, the possible centres are }(5,3)\text{ and }(-5,3).

\displaystyle \text{Case 1: Centre }=(5,3)
\displaystyle \text{The equation of the circle is}
\displaystyle (x-5)^2+(y-3)^2=5^2
\displaystyle \Rightarrow x^2-10x+25+y^2-6y+9=25
\displaystyle \therefore x^2+y^2-10x-6y+9=0.

\displaystyle \text{Case 2: Centre }=(-5,3)
\displaystyle \text{The equation of the circle is}
\displaystyle (x+5)^2+(y-3)^2=5^2
\displaystyle \Rightarrow x^2+10x+25+y^2-6y+9=25
\displaystyle \therefore x^2+y^2+10x-6y+9=0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the equations of the circles passing through two points on the }y\text{-axis}
\displaystyle \text{at distances }3\text{ units from the origin and having radius }5.
\displaystyle \text{Answer:}
\displaystyle \text{The two points on the }y\text{-axis are }(0,3)\text{ and }(0,-3).
\displaystyle \text{Let the centre of the circle be }(h,k).
\displaystyle \text{Since the radius is }5,\text{ its equation is}
\displaystyle (x-h)^2+(y-k)^2=25.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{Since the circle passes through }(0,3),
\displaystyle h^2+(3-k)^2=25
\displaystyle \Rightarrow h^2+k^2-6k=16.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{Since the circle also passes through }(0,-3),
\displaystyle h^2+(-3-k)^2=25
\displaystyle \Rightarrow h^2+k^2+6k=16.\qquad\ldots\ldots\text{(iii)}
\displaystyle \text{Subtracting (ii) from (iii),}
\displaystyle 12k=0
\displaystyle \Rightarrow k=0.
\displaystyle \text{Substituting }k=0\text{ in (ii),}
\displaystyle h^2=16
\displaystyle \Rightarrow h=\pm4.
\displaystyle \therefore \text{The possible centres are }(4,0)\text{ and }(-4,0).
\displaystyle \text{Hence the required equations are}
\displaystyle (x-4)^2+y^2=25
\displaystyle \Rightarrow x^2+y^2-8x-9=0,
\displaystyle \text{or}\quad (x+4)^2+y^2=25
\displaystyle \Rightarrow x^2+y^2+8x-9=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If the lines }2x-3y=5\text{ and }3x-4y=7\text{ are diameters of a circle}
\displaystyle \text{of area }154\text{ square units, obtain the equation of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{The centre is the point of intersection of the two diameters.}
\displaystyle 2x-3y=5\qquad\text{and}\qquad 3x-4y=7.
\displaystyle \text{Solving these equations, we get}
\displaystyle x=1,\qquad y=-1.
\displaystyle \therefore \text{Centre }=(1,-1).
\displaystyle \text{Given, area of the circle }=154\text{ square units.}
\displaystyle \pi r^2=154
\displaystyle \Rightarrow \frac{22}{7}r^2=154
\displaystyle \Rightarrow r^2=49
\displaystyle \Rightarrow r=7.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-1)^2+[y-(-1)]^2=7^2
\displaystyle \Rightarrow (x-1)^2+(y+1)^2=49
\displaystyle \Rightarrow x^2-2x+1+y^2+2y+1=49
\displaystyle \therefore x^2+y^2-2x+2y-47=0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If the line }y=\sqrt{3}x+k\text{ touches the circle }x^2+y^2=16,\text{ find}
\displaystyle \text{the value of }k.
\displaystyle \text{Answer:}
\displaystyle \text{The circle }x^2+y^2=16\text{ has centre }(0,0)\text{ and radius }4.
\displaystyle \text{The equation of the tangent can be written as}
\displaystyle \sqrt{3}x-y+k=0.
\displaystyle \text{Since the line touches the circle, the perpendicular distance of the centre}
\displaystyle \text{from the line is equal to the radius.}
\displaystyle \frac{|\sqrt{3}(0)-0+k|}{\sqrt{(\sqrt{3})^2+(-1)^2}}=4
\displaystyle \Rightarrow \frac{|k|}{2}=4
\displaystyle \Rightarrow |k|=8
\displaystyle \therefore k=\pm8.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the equation of the circle having }(1,-2)\text{ as its centre and passing}
\displaystyle \text{through the intersection of the lines }3x+y=14\text{ and }2x+5y=18.
\displaystyle \text{Answer:}
\displaystyle \text{The point of intersection of the lines }3x+y=14\text{ and }2x+5y=18\text{ is }(4,2).
\displaystyle \text{The centre of the circle is }(1,-2).
\displaystyle \text{Therefore, the radius is}
\displaystyle r=\sqrt{(4-1)^2+[2-(-2)]^2}
\displaystyle =\sqrt{3^2+4^2}=5.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-1)^2+[y-(-2)]^2=5^2
\displaystyle \Rightarrow (x-1)^2+(y+2)^2=25
\displaystyle \Rightarrow x^2-2x+1+y^2+4y+4=25
\displaystyle \therefore x^2+y^2-2x+4y-20=0.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If the lines }3x-4y+4=0\text{ and }6x-8y-7=0\text{ are tangents}
\displaystyle \text{to a circle, find the radius of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{The given lines are}
\displaystyle 3x-4y+4=0\qquad\text{and}\qquad6x-8y-7=0.
\displaystyle \text{Dividing the second equation by }2,
\displaystyle 3x-4y-\frac{7}{2}=0.
\displaystyle \text{The two lines have the same coefficients of }x\text{ and }y,
\displaystyle \therefore \text{they are parallel.}
\displaystyle \text{The distance between the two parallel tangents is the diameter of the circle.}
\displaystyle 2r=\frac{\left|4-\left(-\frac{7}{2}\right)\right|}{\sqrt{3^2+(-4)^2}}
\displaystyle =\frac{\frac{15}{2}}{5}=\frac{3}{2}.
\displaystyle \therefore r=\frac{1}{2}\times\frac{3}{2}=\frac{3}{4}.
\displaystyle \therefore \text{The radius of the circle is }\frac{3}{4}\text{ unit.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Show that the point }(x,y)\text{ given by }x=\frac{2at}{1+t^2}\text{ and}
\displaystyle y=a\left(\frac{1-t^2}{1+t^2}\right)\text{ lies on a circle for all real values of }t\text{ such that}
\displaystyle -1\leq t\leq1,\text{ where }a\text{ is any given real number.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=\frac{2at}{1+t^2}\text{ and }y=a\left(\frac{1-t^2}{1+t^2}\right).
\displaystyle \text{Squaring and adding,}
\displaystyle x^2+y^2=\frac{4a^2t^2}{(1+t^2)^2}+\frac{a^2(1-t^2)^2}{(1+t^2)^2}
\displaystyle =\frac{a^2\left[4t^2+(1-t^2)^2\right]}{(1+t^2)^2}
\displaystyle =\frac{a^2(4t^2+1-2t^2+t^4)}{(1+t^2)^2}
\displaystyle =\frac{a^2(1+2t^2+t^4)}{(1+t^2)^2}
\displaystyle =\frac{a^2(1+t^2)^2}{(1+t^2)^2}
\displaystyle =a^2.
\displaystyle \therefore x^2+y^2=a^2.
\displaystyle \text{Hence, for }a\neq0,\text{ the point }(x,y)\text{ lies on a circle with centre }(0,0)
\displaystyle \text{and radius }|a|\text{ for every }t\text{ such that }-1\leq t\leq1.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The circle }x^2+y^2-2x-2y+1=0\text{ is rolled along the positive}
\displaystyle \text{direction of the }x\text{-axis and makes one complete roll. Find its equation in the new position.}
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the circle is}
\displaystyle x^2+y^2-2x-2y+1=0.
\displaystyle \Rightarrow x^2-2x+1+y^2-2y+1=1
\displaystyle \Rightarrow (x-1)^2+(y-1)^2=1.
\displaystyle \therefore \text{Centre }=(1,1)\text{ and Radius }=1.
\displaystyle \text{The distance covered in one complete roll is equal to the circumference of the circle.}
\displaystyle 2\pi r=2\pi(1)=2\pi.
\displaystyle \text{Therefore, the centre moves }2\pi\text{ units in the positive direction of the }x\text{-axis.}
\displaystyle \therefore \text{The new centre is }(1+2\pi,1).
\displaystyle \text{The radius remains unchanged. Hence, the new equation is}
\displaystyle [x-(1+2\pi)]^2+(y-1)^2=1
\displaystyle \therefore (x-1-2\pi)^2+(y-1)^2=1.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{One diameter of the circle circumscribing the rectangle }ABCD\text{ is}
\displaystyle 4y=x+7.\text{ If the coordinates of }A\text{ and }B\text{ are }(-3,4)\text{ and }(5,4)
\displaystyle \text{respectively, find the equation of the circle.}
\displaystyle \text{Answer:} 2021-02-14_13-53-11
\displaystyle \text{Given, }A(-3,4)\text{ and }B(5,4).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle \left(\frac{-3+5}{2},\frac{4+4}{2}\right)=(1,4).
\displaystyle \text{Since }AB\text{ is horizontal, its perpendicular bisector is}
\displaystyle x=1.
\displaystyle \text{The centre of the circle lies on the perpendicular bisector of }AB
\displaystyle \text{and also on the given diameter }4y=x+7.
\displaystyle \text{Substituting }x=1\text{ in }4y=x+7,
\displaystyle 4y=8
\displaystyle \Rightarrow y=2.
\displaystyle \therefore \text{The centre is }C(1,2).
\displaystyle \text{The radius is}
\displaystyle r=CA=\sqrt{[1-(-3)]^2+(2-4)^2}
\displaystyle =\sqrt{4^2+(-2)^2}=\sqrt{20}=2\sqrt{5}.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-1)^2+(y-2)^2=20
\displaystyle \Rightarrow x^2-2x+1+y^2-4y+4=20
\displaystyle \therefore x^2+y^2-2x-4y-15=0.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If the line }2x-y+1=0\text{ touches a circle at the point }(2,5)\text{ and}
\displaystyle \text{the centre of the circle lies on the line }x+y-9=0,\text{ find its equation.}
\displaystyle \text{Answer:}
\displaystyle \text{The radius through the point of contact is perpendicular to the tangent.}
\displaystyle \text{The slope of the tangent }2x-y+1=0\text{ is }2.
\displaystyle \therefore \text{The slope of the radius through }(2,5)\text{ is }-\frac12.
\displaystyle \text{Hence, the equation of the radius through }(2,5)\text{ is}
\displaystyle y-5=-\frac12(x-2)2021-02-14_13-52-54
\displaystyle \Rightarrow x+2y-12=0.
\displaystyle \text{The centre also lies on }x+y-9=0.
\displaystyle \text{Solving }x+2y-12=0\text{ and }x+y-9=0,
\displaystyle y=3\qquad\text{and}\qquad x=6.
\displaystyle \therefore \text{The centre is }C(6,3).
\displaystyle \text{The radius is}
\displaystyle r=\sqrt{(6-2)^2+(3-5)^2}
\displaystyle =\sqrt{16+4}=\sqrt{20}=2\sqrt5.
\displaystyle \text{Therefore, the equation of the circle is}
\displaystyle (x-6)^2+(y-3)^2=20.
\displaystyle \\

 

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