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The extremities of a diameter of a circle have co ordinates $(6,-2)$ and $(6,8)$. What length does the circle intercept along $X$ axis..?

My Attempt: Let, $(6,-2)\equiv (x_1,y_1)$ and $(6,8)\equiv (x_2,y_2)$ be the end points of diameter of circle. Now, $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$ $$(x-6)(x-6)+(y+2)(y-8)=0$$ $$x^2-12x+36+y^2-6y-16=0$$ $$x^2+y^2-12x-6y+20=0$$

is the required equation of the circle. Now, what should I do then?

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    Find the intercepts on the $x$ axis by letting $y=0$, then their difference. – 2017-02-05
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    @dxiv, Please elaborate, a bit more. I.could not get – 2017-02-05
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    $y=0 \implies x^2-12x+20=0\,$, calculate the roots $x_1, x_2$, then $|x_1-x_2|\,$. – 2017-02-05
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    @dxiv, why is $\abs {x_1-x_2}$ need to be done? – 2017-02-05
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    Because "*length*" usually means a non-negative number. – 2017-02-05
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    @@dxiv, I mean why should the difference be taken? – 2017-02-05
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    `What length does the circle intercept` If the circle crosses the X axis at two points, then I guess the "*length*" in question refers to the distance between them. – 2017-02-05

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Continue with your solution -

The lengths of intercepts made by the circle

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

with X and Y axes are $2\sqrt{g^2−c}$ and $2\sqrt{f^2−c}$ respectively.

Or -

Mid point of (6,-2) and (6,8) is (6,3).

Also radius r = $\frac 12 \sqrt{(6-6)^2 + (8+2)^2}$

= $\frac 12 \sqrt {100}$

= 5

Equation of circle is -

$(x - 6)^2 + (y - 3)^2 = (5)^2$

Put y = 0, you will get two values of x subtract the two intercepts to get the length of the chord.

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    Ok. Thank you for correcting my mistake. – 2017-02-05
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    @@Kanwaljit Singh, How do you get the formula i.e x intercept$=2\sqrt {g^2-c}$? – 2017-02-05
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    I learnt it from here. http://www.math-only-math.com/intercepts-on-the-axes-made-by-a-circle.html – 2017-02-05
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    Hope helps you too. – 2017-02-05
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    Always you write two or three times @ before name. So its not ping the user. Only write single @ before name in comment. – 2017-02-05