For what value of a if any is (x - a)^2 + y^2 = 36 tangential to the line x - y = 4

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The line x - y = 4 is tangential to the circle `(x - a)^2 + y^2 = 36` if the line and the circle have only one point of contact.

x - y = 4 => y = x - 4

Substitute this in `(x - a)^2 + y^2 = 36`

=> `(x - a)^2 + (x - 4)^2 = 36`

=> `x^2 + a^2 - 2ax + x^2 + 16 - 8x = 36`

=> `2x^2 - 2ax - 8x + a^2 - 20 = 0`

This quadratic equation should have a common root. This is the case when `(-2a - 8)^2 - 4*2*(a^2 - 20) = 0`

=> `(-a - 4)^2 - 2*(a^2 - 20) = 0`

=> `a^2 + 16 + 8a - 2a^2 + 40 = 0`

=> `-a^2 + 8a + 56 = 0`

=> `a^2 - 8a - 56 = 0`

a = `(8 +- sqrt(64 + 224))/2`

= `4 +- 6*sqrt 2`

The values of a for which the line x - y = 4 is a tangent of the circle `(x - a)^2 + y^2 = 36` are `4 +- 6*sqrt 2`

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Given equation of circle is

`(x-a)^2+y^2=(6)^2`

line      x - y -4=0  will be tangent to the circle if distance between line and centre of circle equals to radius of the circle.

`+-6=(a-0-4)/sqrt(1^2+(-1)^2)`

`+-6=(a-4)/sqrt(2)`

`a-4=+-6sqrt(2)`

`a=4+-6sqrt(2)`

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