Question

Q.4 Find the equation of circle which passes through the origin and cuts of chords of length 4 and 6 on the positive side of X - axis and y axis respectively.

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Answer 1

$\begin{lgathered}\Large\bf{\color{indigo}GiVeN,} \\\end{lgathered}$

Length of chord on x-axis = 4

Length of chord on y-axis = 6

$\begin{lgathered}\bf\red{We\:know\:that,} \\\end{lgathered}$

☆ General equation of circle,

$\begin{lgathered}\orange\bigstar\:\:\bf\blue{x^2\:+\:y^2\:+\:2gx\:+\:2fy\:+\:c\:=\:0\:} \\\end{lgathered}$

✒ We observe that (0 , 0) , (4 , 0) & (0 , 6) are on circle, i.e. co-ordinates of the circle.

✔ So putting these points in the general equation we have,

For point (0 , 0) ;-

$\begin{lgathered}\longmapsto\:\:\bf{0^2\:+\:0^2\:+\:2g.0\:+\:2f.0\:+\:c\:=\:0\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf{0\:+\:0\:+\:0\:+\:0\:+\:c\:=\:0\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf\pink{c\:=\:0\:} \\\end{lgathered}$

For point (4 , 0) ;-

$\begin{lgathered}\longmapsto\:\:\bf{4^2\:+\:0^2\:+\:2g.4\:+\:2f.0\:+\:0\:=\:0\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf{16\:+\:0\:+\:8g\:+\:0\:+\:0\:=\:0\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf{8g\:=\:-16\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf{g\:=\:\dfrac{-16}{8}\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf\purple{g\:=\:-2\:} \\\end{lgathered}$

For point (0 , 6) ;-

$\begin{lgathered}\longmapsto\:\:\bf{0^2\:+\:6^2\:+\:2g.0\:+\:2f.6\:+\:0\:=\:0\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf{12f\:=\:-36\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf{f\:=\:\dfrac{-36}{12}\:} \\\end{lgathered}$

$\begin{lgathered}\longmapsto\:\:\bf\green{f\:=\:-3\:} \\\end{lgathered}$

☆ Put the value of f , g & c in the equation of circle, we get

$\begin{lgathered}:\implies\:\:\bf{x^2\:+\:y^2\:+\:2.(-2).x\:+\:2.(-3).y\:+\:0\:=\:0\:} \\\end{lgathered}$

$\begin{lgathered}:\implies\:\:\bf{\color{olive}x^2\:+\:y^2\:-\:4x\:-\:6y\:=\:0\:} \\\end{lgathered}$

$\Large\bold\therefore$ The equation of circle is $\bf{\color{coral}x^2\:+\:y^2\:-\:4x\:-\:6y\:=\:0\:}$