Tangents to circles (1 Viewer)

ashokkumar · Feb 19, 2010

The line 4x + 3y - 60 = 0 and both co-ordinate axes are tangents to a circle. Find the radius of the circle.

Drongoski · Feb 19, 2010

ashokkumar said:
The line 4x + 3y - 60 = 0 and both co-ordinate axes are tangents to a circle. Find the radius of the circle.

Very Interesting

I thought this was a straightforward question. Upon further investigation I found the problem to be more interesting than I thought. There are 4 answers as follows. Graph the circles on a large piece of paper to verify. This means the question itself is misleading in that it suggests the circle is unique.

Quad-1: 2 circles possible

i) centre (5,5) radius 5
ii) centre (30,30) radius 30

Quad-2: 1 circle centre (-15,15) radius 15

Quad-3: Not possible

Quad-4: 1 circle: centre (10,-10) radius 10

ashokkumar · Feb 19, 2010

umm.. can you explain how you got that?

Drongoski · Feb 19, 2010

$\100dpi In\ Quadrant-2:\\ \\ we\ have\ circle\ centre (-r,r) \ radius\ r : \\ \\ distance\ of\ centre\ from\ line\ \ 4x+3y-60=0\ \ is\ related\ by:\\ \\ \frac{\left | 4x_{1}+3y_{1}-60 \right |}{\sqrt{4^{2}+3^{2}}}\ =\ r\\ \\ \\ \therefore\ \ \frac{\left | 4\cdot (-r)+3\cdot r-60\left | }{5}\ =\ r\\ \\ \therefore \ \ \left | -r-60 \right |\ =\ 5r\\ \\ \therefore \ \ -r-60\ =\ \pm 5r\\ \\ \therefore \ \ case-1:\ \ -r-60\ =\ 5r\ \ \Rightarrow r=-10\ (impossible\ since\ r> 0)\\ \\ \therefore \ \ case-2:\ \ -r-60\ =\ -5r\ \ \Rightarrow \ r=15\\ \\ \therefore \ \ we\ have\ circle\ centre\ (-15,15)\ and\ radius\ 15$

You can consider the other quadrants similarly.

Where is this question from?

ashokkumar · Feb 20, 2010

My teacher said it was a James Ruse question. lol

Tangents to circles (1 Viewer)

ashokkumar

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Drongoski

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ashokkumar

Mr

Drongoski

Well-Known Member

ashokkumar

Mr

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