
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
610.8k+ 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.
Recently Updated Pages
Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

The coating formed on the metals such as iron silver class 12 chemistry CBSE

Metals are refined by using different methods Which class 12 chemistry CBSE

What do you understand by denaturation of proteins class 12 chemistry CBSE

Assertion Nitrobenzene is used as a solvent in FriedelCrafts class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

