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'Find the equation of the circle with its center at $M(4,3)$ which intersects the circle $(x-3)^2+y^2=5$ perpendicularly'

How can 2 circles have a perpendicular intersection, is this even possible? And if so, how does one tackle these kind of problems?

I think they mean this with perpendicular intersection, I don't know how its called in English precisely (I just translated it literally from my language):

enter image description here

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what is $M(4,3)$? – lab bhattacharjee Oct 23 '12 at 13:10
It's the center – JohnPhteven Oct 23 '12 at 13:10
Two circles intersect each other perpendicularly if the corresponding tangent lines to each circle at the intersection point are perpendicular. Just for curiosity: you aren't allowed to use differential calculus here, are you? – DonAntonio Oct 23 '12 at 13:14
No sir, analytic geometry only. – JohnPhteven Oct 23 '12 at 13:15
up vote 1 down vote accepted

Using the Pythagorean theorem, you want $$r^2+5=d^2$$ where $r$ is the radius of the other circle and $d$ is the distance between the two circles. If the other circle has its centre in $(4, 3)$, the distance is $\sqrt{10}$ and the the equation becomes $r^2=10-5=5$ so the equation for the circle becomes $$(x-4)^2+(y-3)^2=5.$$

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Is this always true for circles intersecting perpendicularly? Should I always use $r^2$(1st circle)+ $r^2$(2nd circle)=$d^2$ – JohnPhteven Oct 23 '12 at 13:39
Yes, draw two intersecting circles and draw a triangle consisting of the radii of the two circles to a point of intersection and a line between the centres. Now the angle of the radii are right only if $r_1^2+r_2^2=d^2$, see However, two other answers seem to get a different result from mine, so I probably have a calculation error somewhere. – Max Morin Oct 23 '12 at 13:45
That is a huge help! – JohnPhteven Oct 23 '12 at 13:48
Ok, now i see. The others are interpreting your question as that you want two circles just touching each other in a single point, while I interpreted it like your picture suggested, hence the different answers. – Max Morin Oct 23 '12 at 13:56

Consider, for example, two circles of radius $1$, one with centre $(0,0)$ and one with centre $(1,1)$. Then at the point $(1,0)$, the tangent lines to the circles are perpendicular - that's what it means.

In fact, by considering translations of the two circles in this example, you could probably turn it into a solution to the problem.

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It's not - that isn't a circle. Speaking strictly in terms of geometry, take a point on a circle, and extend a radius from the centre of the circle to that point. Then there will be a unique perpendicular line which meets the radius at 90 degrees - that is the tangent at that particular point. When the point is an intersection of two circles, you can do this for each circle, and get a tangent line for each circle. If these meet at 90 degrees, then the circles are perpendicular at that point. – user123123 Oct 23 '12 at 13:28

I guess that this means a terminology when two circles are tangents. In this case you find the distance between the point and the center of the circle. \begin{equation} \| (4,3) -(3,0)\| = \sqrt{(4-3)^2 +(3-0)^2} = \sqrt{10}. \end{equation} Then the answer will be \begin{equation} (x-4)^2+(y-3)^2 = (\sqrt{10}-\sqrt{5})^2. \end{equation}

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No, if two circles are tangent they do not intersect perpendicularly. – Ross Millikan Oct 23 '12 at 14:12

For the general case, let the radii of the two intersecting circles be $r_1$ and $r_2$ and the distance between the centers be $d$. If you draw the radii to the point of intersection and the segment between the centers, you have a right triangle, giving $r_1^2+r_2^2=d^2$. You are given $d=\sqrt {10}, r_1=\sqrt 5,$ so $r_2=\sqrt 5$

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We know the equation of the tangent at $P(h,k)$ of $(x-a)^2+(y-b)^2=r^2$ is $(x-h)(h-a)+(y-k)(k-b)=0$, so the gradient is $-\frac{h-a}{k-b}$

and the parametric equation of the circle is $x=a+r\cos A,y=b+r\sin B$, so the gradient at $(a+r\cos A,b+r\sin B)$ is $-\cot A$

For $(x-3)^2+y^2=5,a=3,b=0,r=\sqrt 5,$ so, the gradient at $(3+\sqrt 5\cos B,\sqrt 5\sin B)$ is $-\cot B$

For the required circle $(x-4)^2+(y-3)^2=R^2,a=4,b=3,r=R$ so the gradient at $(4+R\cos C,3+R\sin C)$ is $-\cot C$

To be perpendicular, $-\cot C(-\cot B)=-1$ $\implies \tan B=-\cot C=\tan(\frac{\pi}2+C) \implies B=\frac{\pi}2+C$

So, $(3+\sqrt 5\cos B,\sqrt 5\sin B)$ becomes $(3-\sqrt 5\sin C,\sqrt5\cos C)$

So, $3-\sqrt5\sin C=4+R\cos C$,$\sqrt5\cos C=3+R\sin C$

or $R\cos C+\sqrt5 \sin C=-1$ and $R\sin C-\sqrt 5cos C=-3$

Squaring and adding $R^2+ 5=1^2+3^2\implies R=\sqrt5$ as $R>0$

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This is not correct, as shown by the two disagreeing and simpler answers. – Ross Millikan Oct 23 '12 at 14:13
@RossMillikan, I've rectified the mistake in the calculation. Acceptably, there are much simpler ways to calculate it. – lab bhattacharjee Oct 23 '12 at 14:30

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