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Bunuel - the graph in post above is for a circle at centre (0,8) and not (8,0) as asked in the question

one can use coordinate geometry formula to solve this as well.

Since the line y=x is tangent to the circle with centre (8,0) the length of the perpendicular dropped from point (8,0) to this line would be equal to radius.

Now, length of perpendicular dropped from point (X,Y) to any line ax+by+c=0 is given by the formula Mod ((aX+bY+c)/(a^2+b^2)^0.5))

So, length of perpendicular from point (8,0) is 8/(2)^0.5

and Diameter = 2*radius = 16/(2)^0.5 = 8*(2)^0.5. Answer C
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Bunuel - the graph in post above is for a circle at centre (0,8) and not (8,0) as asked in the question

Yes, actually it is. Though as the diagram is symmetric to y=x it doesn't matter at all: for any point (8,0), (0,8), (-8,0), (0,-8) the answer would be the same.

Also I wouldn't recommend approach you used as it involves memorizing the formula you don't really need for the GMAT. As you can see one can solve this problem in pretty straightforward way without this formula.
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Yes - symmetry makes it immaterial in this particular case.

I agree with you about not needing to remember this formula, but if one knows this concept, quite a few questions of similar nature can be done faster - so this might be useful for some people.
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Yes - symmetry makes it immaterial in this particular case.

I agree with you about not needing to remember this formula, but if one knows this concept, quite a few questions of similar nature can be done faster - so this might be useful for some people.

OK then here it is:

DISTANCE BETWEEN THE LINE AND POINT:
Line: \(ay+bx+c=0\), point \((x_1,y_1)\)

\(d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}\)

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is \((0,0)\) -->

\(d=\frac{|c|}{\sqrt{a^2+b^2}}\)

Please note again that it's highly unlikely that you'll need this on the GMAT.

Problems:
distance-to-line-point-on-coordinate-plane-please-help-102776.html
gmatclub-hardest-questions-co-geometry-q1-92566.html
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Hi bunuel please explain the solution in more steps, it is unclear to me. From where does square root 2 came? please help me to understand this.
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Hi bunuel please explain the solution in more steps, it is unclear to me. From where does square root 2 came? please help me to understand this.

Triangle formed is a 45-45-90 right triangle.

• A right triangle where the angles are 45°, 45°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.

For more on this issues check Triangles chapter of Math Book.
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