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Circle(s) touching x-axis at a distance ...

Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length `2sqrt7` on y-axis is (are)

A

`x^(2)+y^(2)-6x+8y+9=0`

B

`x^(2)+y^(2)-6x+7y+9=0`

C

`x^(2)+y^(2)-6x-8y+9=0`

D

`x^(2)+y^(2)-6x-7y+9=0`

Text Solution

Verified by Experts

The correct Answer is:
A, C
PLAN

Hence, the length of intercept on Y-axis s `rArr 2sqrt(f^(2)-c` and if circle touches X-axis
`rArr g^(2) = c`
for `x^(2)+y^(2)+2gx + 2fy+c =0`
Here, `x^(2) + y^(2) +2gx + 2fy + c = 0 `

passes through (3, 0).
`rArr 9 + 6g + c =0` ...(i)
`g^(2)=c` ...(ii)
and `2sqrt(f^(2) -c)=2sqrt7`
` f^(2) -c = 7 ` ... (iii)
From Eqs. (i) and (ii) , we get
`g^(2)+6g+9=0rArr(g+3)^(2)=0`
`rArr` g = -3 and c = 9
` therefore f^(2)=16 rArr f = pm 4`
`therefore x^(2) + y^(2) - 6x pm 8y + 9 = 0`
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Knowledge Check

  • Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 2sqrt(7) on y-axis is (are)

    A
    `x^(2)+y^(2)-6x+8y+9=0`
    B
    `x^(2)+y^(2)-6x+7y+9=0`
    C
    `x^(2)+y^(2)-6x-8y+9=0`
    D
    `x^(2)+y^(2)-6x-7y+9=0`

  • Equation of a circle which touches y-axis at a distance + 4 from origin and intercepts a length 6 on x-axis will be

    A
    `x^(2)+y^(2)+10x -8y +10=0`
    B
    `x^(2)+y^(2) pm 10 x-8y +16=0`
    C
    `x^(2)+y^(2) +10x pm 8y +16=0`
    D
    none

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