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Find the equation of circle whose centre is (3,-1) and which cuts off an intercept of...

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shankcool | Student, Grade 11 | (Level 2) eNoter

Posted June 20, 2012 at 11:46 AM via web

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Find the equation of circle whose centre is (3,-1) and which cuts off an intercept of length 6 from the line 2x - 5y + 18 = 0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 20, 2012 at 1:07 PM (Answer #1)

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You should remember the equation of circle in standard form such that:

`(x-h)^2 + (y-k)^2 = r^2`

The problem provides the coordinates (h,k) of center of circle, hence, you should substitute 3 for h and -1 for k in equation above such that:

`(x-3)^2 + (y+1)^2 = r^2`

Notice that the radius is not given but you may find it using the information provided by the problem.

The line that has the equation `2x - 5y + 18 = 0`  and the length of 6 represents a chord of circle.

You may drop a perpendicular from the center of circle to this chord. The perpendicular line drops in the middle of this chord.

You may find the length of perpendicular using the formula:

`d = |a*x_C + b*y_C + c|/(sqrt(a^2+b^2))`

Notice that ax+by+c represents the equation of chord and `x_C,y_C` , coordinates of the center.

`d = |6 +5 + 18|/(sqrt(4+25))`

`d = |29|/(sqrt(29)) =gt d = sqrt29`

Notice that the radius of circle represents the hypotenuse of right angle triangle that has the lengths of the legs `sqrt29`  and 3.

Using Pythagorean theorem yields:

`r^2 = 29 + 9`

`r^2 = 38`

You may now write the equation of the circle such that:

`(x-3)^2 + (y+1)^2 = 38`

Hence, evaluating the equation of circle under given conditions yields `(x-3)^2 + (y+1)^2 = 38.`  

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