ISBtarget wrote:
I am too weak in this section, could you please explain...what is this formula and why did you do D-1?
please help
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
Now we can do this by finding the equation of a line perpendicular to given line \(y=\frac{3}{4}*x-3\) (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.
There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.
We know the formula to calculate the distance between two points \((x1,y1)\) and \((x2,y2)\): \(d=\sqrt{(x1-x2)^2+(y1-y2)^2}\) BUT there is a formula to calculate the distance between the point (in our case origin) and the line:
DISTANCE BETWEEN THE LINE AND POINT:Line: \(ay+bx+c=0\), point \((x1,y1)\)
\(d=\frac{|ay1+bx1+c|}{\sqrt{a^2+b^2}}\)
DISTANCE BETWEEN THE LINE AND ORIGIN:As origin is \((0,0)\) -->
\(d=\frac{|c|}{\sqrt{a^2+b^2}}\)
So in our case it would be: \(d=\frac{|-3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4\)
So the shortest distance would be: \(2.4-1(radius)=1.4\)
Answer: A.
P.S. Also note that when we have \(x^2+y^2=k\), we have circle (as we have \(x^2\) and \(y^2\)), it's centered at the origin (as coefficients of \(x\) and \(y\) are \(1\)) and the radius of that circle \(r=\sqrt{k}\).
You can check the link of Coordinate Geometry below for more.
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