Circles
3.0 Intercepts made by a circle on coordinate axis
3.0 Intercepts made by a circle on coordinate axis
Let us assume the general form of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$
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| Length of intercept on $X$-axis, $AB = \left| {{x_2} - {x_1}} \right|$ | Length of intercept on $Y$-axis, $CD = \left| {{y_2} - {y_1}} \right|$ |
| The circle intersects on $X$-axis, $y=0$. Therefore, the equation of circle becomes $${x^2} + 2gx + c = 0$$ Roots of equation are ${x_1}$ and ${x_2}$ such that, ${x_1} + {x_2} = - 2g$ and ${x_1} \times {x_2} = c$. | The circle intersects on $Y$-axis, $x=0$. Therefore, the equation of circle becomes $${y^2} + 2fy + c = 0$$ Roots of equation are ${y_1}$ and ${y_2}$ such that, ${y_1} + {y_2} = - 2f$ and ${y_1} \times {y_2} = c$. |
| $X$-intercept ${x_2} - {x_1} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2}} $ $ = \sqrt {{{\left( {{x_2} + {x_1}} \right)}^2} - 4{x_1}{x_2}} $ $ = \sqrt {4{g^2} - 4c} $ $ = 2\sqrt {{g^2} - c} $ | $Y$-intercept ${y_2} - {y_1} = \sqrt {{{\left( {{y_2} - {y_1}} \right)}^2}} $ $ = \sqrt {{{\left( {{y_2} + {y_1}} \right)}^2} - 4{y_1}{y_2}} $ $ = \sqrt {4{f^2} - 4c} $ $ = 2\sqrt {{f^2} - c} $ |
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Question 7. Find the equation of the circle passes through origin $(0,0)$ and has intercepts $2\alpha $ and $2\beta $ on the $X$-axis and $Y$-axis respectively.
Solution: As shown in figure $11$, $OA = 2\alpha $ and $OB = 2\beta $
Therefore, $OM=\alpha $ and $ON = \beta $.
Let us assume the centre of the circle is $C(\alpha ,\beta )$ and radius of circle is $OC = \sqrt {{\alpha ^2} + {\beta ^2}} $
then the equation of circle is $${\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {\alpha ^2} + {\beta ^2}$$ $${x^2} + {y^2} - 2\alpha x - 2\beta y = 0$$
NOTE: If a circle is passing through origin, then constant term is absent i.e., ${x^2} + {y^2} + 2gx + 2fy = 0$.
Question 8. Find the equation of the circle which touches the $X$-axis.
Solution: Let $C(\alpha ,\beta )$ be the centre of the circle, then radius $r = \left| \beta \right|$.
The equation of circle is $${\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {\beta ^2}$$$${x^2} + {y^2} - 2\alpha x - 2\beta y + {\alpha ^2} = 0$$
NOTE: If the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ touches the $X$-axis, then radius$= \sqrt {{g^2} + {f^2} - c} $
or, $\left| { - f} \right| = \sqrt {{g^2} + {f^2} - c} $
or, $c = {g^2}$
Similiarly, if a circle touches $Y$-axis, then $c = {f^2}$.
Question 9. Find the length of intercept made by the circle ${x^2} + {y^2} + 10x - 6y + 9 = 0$ on $X$-axis and $Y$-axis.
Solution: On comparing the given equation of circle with the general form of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get, $g=5,\ f=-3\ and\ c=9$.
Length of intercept on $X$-axis is $\left| {{x_2} - {x_1}} \right| = 2\sqrt {{g^2} - c} = 2\sqrt {{5^2} - 9} = 2\sqrt {25 - 9} = 2 \times 4 = 8$
Length of intercept on $Y$-axis is $\left| {{y_2} - {y_1}} \right| = 2\sqrt {{f^2} - c} = 2\sqrt { - {3^2} - 9} = 0$
Question 10. The circle ${x^2} + {y^2} - 6x - 10y + k = 0$ neither touches nor intersect the coordinate axis. Find the range of $k$.
Solution: Comparing the given equation of circle with the general form of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get, $g=-3\ f=-5\ and\ c=k$.
If circle is not touching the $X$-axis, then $D<0$ i.e., ${b^2} - 4ac < 0$ or $4{g^2} - 4c < 0$
$4 \times - {3^2} - 4k < 0$
or, $k>9...(1)$
If circle is not touching the $Y$-axis, then $D<0$ i.e., ${b^2} - 4ac < 0$ or $4{f^2} - 4c < 0$
$4 \times - {5^2} - 4k < 0$
or, $k>25...(2)$
From equations $(1)$ and $(2)$, we can say that range of $k$ is $(25,\infty )$.
