find the equations of a circle touching the x-axis and passing through the points (7,-2) & (0, -9).

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember the equation of circle in general form such that:

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

The problem provides the information that the circle passes through the point (7,-2), hence you should substitute 7 for x and -2 for y in equation of circle such that:

`(7-h)^2 + (-2-k)^2 = r^2`

The problem provides the information that the circle passes through the point (0,-9), hence you should substitute 0 for x and -9 for y in equation of circle such that:

`(-h)^2 + (-9-k)^2 = r^2`

The problem provides the information that the circle is tangent to x axis, hence the radius of circle is equal to y coordinate of center of circle, hence you need to substitute k for r in equation of circle such that:

`(7-h)^2 + (-2-k)^2 = k^2 `

`(-h)^2 + (-9-k)^2 = k^2`

You need to expand binomials such that:

`49 - 14h + h^2 + 4 + 4k + k^2 = k^2`

`53 - 14h + 4k + h^2 = 0`

`h^2 + 81 + 18k + k^2 = k^2`

`h^2 + 81 + 18k = 0 `

You need to set the equations `53 - 14h + 4k + h^2 = 0`  and `h^2 + 81 + 18k = 0`  equal such that:

`53 - 14h + 4k + h^2 = 81 + 18k + h^2`

`53 - 14h + 4k = 81 + 18k`

`28 + 14k + 14h = 0`

`2 + k + h = 0 =gt k = -h-2`

You need to substitute `-h-2`  for k in `h^2 + 81 + 18k = 0 ` such that:

`h^2 + 81 + 18(-h-2) = 0`

You need to open the brackets such that:

`h^2 + 81 - 18h - 36 = 0`

`h^2 - 18h + 45 = 0`

You should use quadratic formula such that:

`h_(1,2) = (18+-sqrt(324 - 180))/2`

`h_(1,2) = (18+-sqrt144)/2`

`h_(1,2) = (18+-12)/2`

`h_1 = 15 ; h_2 = 3`

Hence, you need to substitute 15 for h in equation `k = -h-2`  such that:

`k = -15-2 = -17`

Hence, you need to substitute 3 for h in equation `k = -h-2`  such that:

`k = -3-2 = -5`

Hence, evaluating the equation of the circle under given conditions yields that there are two possible equations such that:

`(x - 3)^2 + (y + 5)^2 = 5^2`

`(x - 15)^2 + (y + 17)^2 = 17^2`

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malagala | College Teacher | (Level 2) Adjunct Educator

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if the cordinates of the center of the circle is given by (h,k) and if the radius of the circle is given by r, the equation of the circle can be written as,

(x-h)^2 + (y-k)^2 = r^2 ......(1)

as this circle touches x-axis it should pass through the point (h,0) where it touches the x-axis. At the same time it is clear that the radius (r) is equal to the value k as the circle is touching the x- axis.

r = k

Hence the equation (1) can be modified as follows,

(x-h)^2 + (y-k)^2 = k^2 ......(2)

it is also given that the circle passes through the points (7,-2) and (0,-9).

hence these two points should satisfy the equation (2)

Using the point (7,-2) 

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

=> (7-h)^2 + (-2-k)^2 = k^2

=> (49 - 14h +h^2) + (4 +4k +k^2) = k^2

=> h^2 -14h +4K = -53 ...... (3)

 

Using the point (0,-9)


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

=> (0-h)^2 + (-9-k)^2 = k^2

=> (h^2) +( 81 +18k +k^2) = k^2

=> h^2 + 18K = -81 ..... (4)

 

Now we have obtained two equations, (3) and (4) and by solving them we can find the values for h and k

h^2 -14h +4K = -53 ...... (3)

h^2 + 18K = -81 ..... (4)

 

multiply the equation (4) by 2 and multiply equation (3) by 9 and substract each other

9x(3) - 2x(4)

=> 9(h^2) - 126h + 36k - 2(h^2) - 36k = -477 +162

=> 7(h^2) - 126 h = -315 

=>h^2 - 18h =- 45 (divided by 7)

=>h^2 -18h +45 = 0

=> (h -15)(h-3) = 0

in order to satisfy this condition,

h = 15 0r h = 3,

Consider equation (4)

h^2 + 18K = -81 ..... (4)

when h = 15

15^2 +18 k = -81

=> 225 +18k = -81

=> 18 k = -306

=> k = -17

when h = 3

3^2 +18 k = -81

=> 9 +18k = -81

=> 18 k = -90

=> k = -5

therefore we have obtained two centers for the circle, (15,-17) and (3,-5)

That means there are two circles which satisfy the given conditions.

So now lets find out the equations of those equations,

Consider the point (15,-17) where (h =15, k= -17, r = k =17)

(r is a positive value)

the equation can be given by,

(x-h)^2 + (y-k)^2 = k^2 ......(2)

=> (x -15)^2 + (y - (-17))^2 = 17^2

=> (x^2) - 30x +225 + (y^2) + 34y +289 = 289

=> (x^2) + (y^2) -30x +34y +225 = 0


Consider the point (3,-5) where (h =3, k= -5, r = k =5)

(r is a positive value)

the equation can be given by,

(x-h)^2 + (y-k)^2 = k^2 ......(2)

=> (x -3)^2 + (y - (-5))^2 = 5^2

=> (x^2) - 6x +9 + (y^2) + 10y +25 = 25

=> (x^2) + (y^2) -6x +10y +9 = 0


so the equations of the circles are,

(x^2) + (y^2) -6x +10y +9 = 0

 

and

(x^2) + (y^2) -30x +34y +225 = 0

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