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

EXPLANATION.

Equation of circle passes through origin and cuts of chord of length 4 and 6 cm on positive x-axis and y-axis.

General equation of circle = x² + y² + 2gx + 2fy + c = 0.

Co-ordinate of the circle = ( 0,0) , (4,0) , ( 0,6).

put the value of co-ordinates in general equation of circle.

put the co-ordinate = ( 0,0).

⇒ (0)² + (0)² + 2g(0) + 2f(0) + c = 0.

⇒ c = 0.   ........(1).

put the co-ordinate = ( 4,0).

⇒ (4)² + (0)² + 2g(4) + 2f(0) + c = 0.

⇒ 16 + 0 + 8g + 0 + c = 0.

⇒ 16 + 8g + c = 0 ......(2).

put the co-ordinate = (0,6).

⇒ (0)² + (6)² + 2g(0) + 2f(6) + c = 0.

⇒ 0 + 36 + 0 + 12f + c = 0.

⇒ 36 + 12f + c = 0 ....(3).

From equation (1) , (2) and (3) we get,

put the value of equation (1) in equation (2).

⇒ 16 + 8g + 0 = 0.

⇒ 8g = -16.

⇒ g = -2

Again put the value of equation (1) in (3).

⇒ 36 + 12g + 0 = 0.

⇒ 12f = -36.

⇒ f = -3.

Value of ⇒ c = 0, g = -2 , f = -3.

put the value in general equation of circle,

⇒ x² + y² + 2(-2)x + 2(-3)y + 0 = 0.

⇒ x² + y² - 4x - 6y = 0.

Answer 2

$\Large\bf{\color{indigo}GiVeN,}$

• Length of chord on x-axis = 4

• Length of chord on y-axis = 6

$\bf\red{We\:know\:that,}$

☆ General equation of circle,

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

✒ 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) ;-

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

$\longmapsto\:\:\bf{0\:+\:0\:+\:0\:+\:0\:+\:c\:=\:0\:}$

$\longmapsto\:\:\bf\pink{c\:=\:0\:}$

For point (4 , 0) ;-

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

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

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

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

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

For point (0 , 6) ;-

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

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

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

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

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

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

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

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