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Question: A circle touches x-axis at (2, 0) and also the line y=x in the first quadrant then its radius is (...

A circle touches x-axis at (2, 0) and also the line y=x in the first quadrant then its radius is
(a) 2−1\sqrt{2}-1
(b) 2−22-\sqrt{2}
(c) 2(2−1)2\left( \sqrt{2}-1 \right)
(d) 2+1\sqrt{2}+1

Explanation

Solution

Hint : First, we need to assume that the distance in the y axis of the centre of the circle is k and we need to draw the figure of the circle which touches the point (2, 0) and also touches the line y=x in first quadrant. Then, by using the figure we can clearly see that the distance marked as r in the circle for both the radius is equal. Then, by using the distance formula between two points with coordinates (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) as (x2−x1)2+(y2−y1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} , we get the value of radius of the circle.

Complete step-by-step answer :
In this question, we are supposed to find the radius of the circle which touches the x-axis at (2, 0) and also touches the line y=x in the first quadrant.
Firstly, we need to draw the figure of the circle which touches the point (2, 0) and also touches the line y=x in the first quadrant.
Before that, we need to assume that the distance in the y axis of the centre of the circle is k.
So, the centre of the circle is as (2, k) and r is the radius of the circle as shown in the diagram below:

Now, by using the figure we can clearly see that the distance marked as r in the circle for both the radius is equal.
Then by using this condition and equating the value of both radiuses as equal, we get:
∣2−k1+1∣=(2−2)2+(k−0)2\left| \dfrac{2-k}{\sqrt{1+1}} \right|=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( k-0 \right)}^{2}}}
So, the above condition is found by using the distance formula between two points with coordinates (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) as:
(x2−x1)2+(y2−y1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}
Now, by solving the expression, we get:
∣2−k2∣=0+(k)2 ⇒∣2−k2∣=k \begin{aligned} & \left| \dfrac{2-k}{\sqrt{2}} \right|=\sqrt{0+{{\left( k \right)}^{2}}} \\\ & \Rightarrow \left| \dfrac{2-k}{\sqrt{2}} \right|=k \\\ \end{aligned}
Now, by taking the positive value of the modulus only to get the value of k as:
2−k=2k ⇒2=2k+k ⇒2=k(2+1) ⇒k=22+1 \begin{aligned} & 2-k=\sqrt{2}k \\\ & \Rightarrow 2=\sqrt{2}k+k \\\ & \Rightarrow 2=k\left( \sqrt{2}+1 \right) \\\ & \Rightarrow k=\dfrac{2}{\sqrt{2}+1} \\\ \end{aligned}
Then, by rationalising the above value of k to get a value that matches any of the options as:
k=22+1×2−12−1 ⇒k=2(2−1)(2)2−12 ⇒k=2(2−1)2−1 ⇒k=2(2−1) \begin{aligned} & k=\dfrac{2}{\sqrt{2}+1}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1} \\\ & \Rightarrow k=\dfrac{2\left( \sqrt{2}-1 \right)}{{{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}} \\\ & \Rightarrow k=\dfrac{2\left( \sqrt{2}-1 \right)}{2-1} \\\ & \Rightarrow k=2\left( \sqrt{2}-1 \right) \\\ \end{aligned}
So, we can see that the radius circle is 2(2−1)2\left( \sqrt{2}-1 \right) which is also the value of k.
So, the correct answer is “Option C”.

Note : Now, to solve these type of the questions we need to know some of the basic formulas to calculate the distance between the two points let AB be the line with coordinates as A (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and B (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) . Then, the distance between the points AB is given by:
AB‾=(x2−x1)2+(y2−y1)2\overline{AB}=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}. We equate both the equations here as those represent the radii of the same circle.