NCERT Solutions · Class 11 Mathematics Conic Sections

Locating the centre from equal radii. Let the centre be \(\displaystyle C(h,k)\). It lies on \(\displaystyle x-3y-11=0\), so \[h=3k+11. \tag{1}\]\(\displaystyle C\) is equidistant from \(\displaystyle A(2,3)\) and \(\displaystyle B(-1,1)\): \[(h-2)^{2}+(k-3)^{2}=(h+1)^{2}+(k-1)^{2}\] \[-4h+4-6k+9=2h+1-2k+1\] \[6h+4k=11. \tag{2}\]Substituting ($\displaystyle 1$) into ($\displaystyle 2$): \(\displaystyle 6(3k+11)+4k=11\Rightarrow 22k=-55\Rightarrow k=-\dfrac52\), so \(\displaystyle h=3\left(-\dfrac52\right)+11=\dfrac72\).Radius: \(\displaystyle r^{2}=\left(\dfrac72-2\right)^{2}+\left(-\dfrac52-3\right)^{2}=\dfrac94+\dfrac{121}{4}=\dfrac{65}{2}\) (check with \(\displaystyle B\): \(\displaystyle \left(\dfrac72+1\right)^{2}+\left(-\dfrac52-1\right)^{2}=\dfrac{81}{4}+\dfrac{49}{4}=\dfrac{65}{2}\) ✓).\[\left(x-\frac72\right)^{2}+\left(y+\frac52\right)^{2}=\frac{65}{2}\] \[x^{2}-7x+\frac{49}{4}+y^{2}+5y+\frac{25}{4}=\frac{130}{4}\]\(\displaystyle x^{2}+y^{2}-7x+5y-14=0\)

Centre on the \(\displaystyle x\)-axis. A point of the \(\displaystyle x\)-axis has \(\displaystyle y=0\), so let the centre be \(\displaystyle C(h,0)\). The circle has radius \(\displaystyle 5\) and passes through \(\displaystyle (2,3)\), so the distance from \(\displaystyle C\) to \(\displaystyle (2,3)\) is \(\displaystyle 5\): \[(2-h)^{2}+(3-0)^{2}=25\] \[(2-h)^{2}=16\ \Longrightarrow\ 2-h=\pm4.\]So \(\displaystyle h=-2\) or \(\displaystyle h=6\); both are admissible, giving two circles: \[(x+2)^{2}+y^{2}=25\qquad\text{and}\qquad (x-6)^{2}+y^{2}=25.\]\(\displaystyle x^{2}+y^{2}+4x-21=0\) or \(\displaystyle x^{2}+y^{2}-12x+11=0\)

General equation, three conditions. Making intercepts \(\displaystyle a\) and \(\displaystyle b\) on the axes means the circle cuts the \(\displaystyle x\)-axis at \(\displaystyle (a,0)\) and the \(\displaystyle y\)-axis at \(\displaystyle (0,b)\). Take the general circle \[x^{2}+y^{2}+2gx+2fy+c=0.\]Through \(\displaystyle (0,0)\): \(\displaystyle c=0\).Through \(\displaystyle (a,0)\): \(\displaystyle a^{2}+2ga=0\Rightarrow g=-\dfrac{a}{2}\) (as \(\displaystyle a\neq0\)).Through \(\displaystyle (0,b)\): \(\displaystyle b^{2}+2fb=0\Rightarrow f=-\dfrac{b}{2}\) (as \(\displaystyle b\neq0\)).Hence \(\displaystyle x^{2}+y^{2}-ax-by=0\). All three points satisfy it, so it is the required circle.\(\displaystyle x^{2}+y^{2}-ax-by=0\)

Radius as a distance. The radius is the distance from the centre \(\displaystyle (2,2)\) to the point \(\displaystyle (4,5)\) on the circle: \[r=\sqrt{(4-2)^{2}+(5-2)^{2}}=\sqrt{4+9}=\sqrt{13}.\]Using \(\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}\): \[(x-2)^{2}+(y-2)^{2}=13\] \[x^{2}-4x+4+y^{2}-4y+4=13\]\(\displaystyle x^{2}+y^{2}-4x-4y-5=0\)

Compare the distance from the centre with the radius. The circle \(\displaystyle x^{2}+y^{2}=25\) has centre \(\displaystyle (0,0)\) and radius \(\displaystyle 5\). A point lies inside, on, or outside according as its distance from the centre is less than, equal to, or greater than \(\displaystyle 5\) — equivalently, according as \(\displaystyle x^{2}+y^{2}\) is less than, equal to, or greater than \(\displaystyle 25\).For \(\displaystyle (-2.5,\,3.5)\): \[x^{2}+y^{2}=(-2.5)^{2}+(3.5)^{2}=6.25+12.25=18.5<25.\](The distance is \(\displaystyle \sqrt{18.5}\approx4.30<5\).)The point lies inside the circle.