Show that the circles \[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}}\] and \[{{(x-b)}^{2}}+{{(y-a)}^{2}}={{c}^{2}}\] touch each other if \[a=b\pm \sqrt{2}c\].
Answer
651.6k+ views
Hint: We will compare both the equation of circles with the general form to get the center and radius of both the circles. Then we will use the distance formula \[AB=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] between two points A and B.
Complete step-by-step answer:
We know that the general equation of the circle is \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}.........(1)\] and center is (h, k) and r is the radius.
Now in the question the equation of first circle is \[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}}.......(2)\]. So now comparing equation (2) with equation (1) we get center as (a, b) and radius as c.
Now in the question the equation of second circle is \[{{(x-b)}^{2}}+{{(y-a)}^{2}}={{c}^{2}}.......(3)\]. So now comparing equation (3) with equation (1) we get center as (b, a) and radius as c.
Also it is given that the two circles touch each other. Also the radius is same in both the circles. So joining the two centers of the circles with coordinates (a, b) and (b, a). From this we can say that the total distance between these two coordinates is 2c. Hence applying the distance formula we get, \[\sqrt{{{(a-b)}^{2}}+{{(b-a)}^{2}}}=2c.......(4)\]
Now squaring both sides of equation (4) we get,
\[\Rightarrow {{(a-b)}^{2}}+{{(b-a)}^{2}}=4{{c}^{2}}.......(5)\]
Now expanding the squares in the left hand side of the equation (5) we get,
\[\Rightarrow {{a}^{2}}+{{b}^{2}}-2ab+{{b}^{2}}+{{a}^{2}}-2ab=4{{c}^{2}}.......(6)\]
Adding all the similar terms in equation (6) and rearranging we get,
\[\Rightarrow 2({{a}^{2}}+{{b}^{2}}-2ab)=4{{c}^{2}}.......(7)\]
Now simplifying in equation (7) we get,
\[\Rightarrow {{(a-b)}^{2}}=2{{c}^{2}}.......(8)\]
Now taking square root on both sides of equation (8) we get,
\[\Rightarrow a-b=\pm \sqrt{2}c.......(9)\]
Now isolating a in equation (9) we get,
\[\Rightarrow a=b\pm \sqrt{2}c\]. Hence proved.
Note: Here remembering the general formula of circle and related properties is the key. Also knowing the distance formula between two points is important. We in a hurry can make a mistake in equation (9) by only writing + and hence we need to be careful while doing this step.
Complete step-by-step answer:
We know that the general equation of the circle is \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}.........(1)\] and center is (h, k) and r is the radius.
Now in the question the equation of first circle is \[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}}.......(2)\]. So now comparing equation (2) with equation (1) we get center as (a, b) and radius as c.
Now in the question the equation of second circle is \[{{(x-b)}^{2}}+{{(y-a)}^{2}}={{c}^{2}}.......(3)\]. So now comparing equation (3) with equation (1) we get center as (b, a) and radius as c.
Also it is given that the two circles touch each other. Also the radius is same in both the circles. So joining the two centers of the circles with coordinates (a, b) and (b, a). From this we can say that the total distance between these two coordinates is 2c. Hence applying the distance formula we get, \[\sqrt{{{(a-b)}^{2}}+{{(b-a)}^{2}}}=2c.......(4)\]
Now squaring both sides of equation (4) we get,
\[\Rightarrow {{(a-b)}^{2}}+{{(b-a)}^{2}}=4{{c}^{2}}.......(5)\]
Now expanding the squares in the left hand side of the equation (5) we get,
\[\Rightarrow {{a}^{2}}+{{b}^{2}}-2ab+{{b}^{2}}+{{a}^{2}}-2ab=4{{c}^{2}}.......(6)\]
Adding all the similar terms in equation (6) and rearranging we get,
\[\Rightarrow 2({{a}^{2}}+{{b}^{2}}-2ab)=4{{c}^{2}}.......(7)\]
Now simplifying in equation (7) we get,
\[\Rightarrow {{(a-b)}^{2}}=2{{c}^{2}}.......(8)\]
Now taking square root on both sides of equation (8) we get,
\[\Rightarrow a-b=\pm \sqrt{2}c.......(9)\]
Now isolating a in equation (9) we get,
\[\Rightarrow a=b\pm \sqrt{2}c\]. Hence proved.
Note: Here remembering the general formula of circle and related properties is the key. Also knowing the distance formula between two points is important. We in a hurry can make a mistake in equation (9) by only writing + and hence we need to be careful while doing this step.
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