Math, asked by srush8, 2 months ago

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.​

Answers

Answered by Madankumar808103
3

Answer:

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.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.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.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.

Answered by palsabita1957
7

GiVeN,

Length of chord on x-axis = 4

Length of chord on y-axis = 6

We know that,

General equation of circle,

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) :-

(0)^2+(0)^2 +2g×0+2f×0+c=0

⟼0+0+0+0+c=0

⟼c=0

For point (4 , 0)

(4)^2+(0)^2+2g.4+2f.0+0=0

⟼16+0+8g+0+0=0

⟼8g=−16

⟼g= −16/8

⟼g=−2

☆For point (0 , 6) ;-

(0)^2+(6)^2 +2g.0+2f.6+0=0

⟼12f=−36

⟼f=−36/12

⟼f=−3

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

⟹x^2 +y^2+2.(−2).x+2.(−3).y+0=0

⟹x^2 +y^2 −4x−6y=0

⟹ x^2 +y^2 −4x−6y=0.

The equation of circle is x^2 +y^2−4x−6y=0.

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