Solveeit Logo
Login

Question

Question: Find the length of the diameter of the circle which passes through the point \((2,3)\) and touches t...

Find the length of the diameter of the circle which passes through the point (2,3)(2,3) and touches the x axis at the point (1,0)(1,0). Choose from the correct answer.
(A) 65\dfrac{6}{5}
(B) 105\dfrac{{10}}{5}
(C) 53\dfrac{5}{3}
(D) 103\dfrac{{10}}{3}

Explanation

Solution

In this question we use the equation of the circle which is (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}, here h and k are the x and y coordinates of the center of the circle and r is the radius. By using this equation we find the value of the radius by putting the values in the equation given in the question. As diameter = 2 times of the radius. With the help of the value of radius we find the value of diameter.

Complete step-by-step answer:
According to the question we have to find the value of diameter of a circle which touches the x axis at (1,0)(1,0) and passes through (2,3)(2,3).
Now let us consider the coordinates of the center of the circle be (h,k) and r be the radius of the circle
For finding the value of diameter we have to find the value of radius first
∵Diameter=2×radius\because Diameter = 2 \times radius
For finding the value radius we use the equation of the circle
(x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
As given in the question the circle touches x axis at (1,0)(1,0),
∴\therefore Radius of the circle r = k
Now, the equation of the circle is
(x−h)2+(y−k)2=k2{(x - h)^2} + {(y - k)^2} = {k^2}
Now the circle passes through points (1,0)(1,0)and (2,3)(2,3), thus these both coordinates will satisfy the equation of the circle.
Putting (1,0)(1,0) in the equation we get,
∴\therefore (1−h)2+(0−k)2=k2{(1 - h)^2} + {(0 - k)^2} = {k^2}
⇒(1−h)2+(−k)2=k2\Rightarrow {(1 - h)^2} + {( - k)^2} = {k^2}
By solving the equation, we get,
⇒(1−h)2+k2=k2\Rightarrow {(1 - h)^2} + {k^2} = {k^2}
⇒(1−h)2=k2−k2 ⇒(1−h)2=0  \Rightarrow {(1 - h)^2} = {k^2} - {k^2} \\\ \Rightarrow {(1 - h)^2} = 0 \\\
By solving the equation, we get the value of h,
⇒h2=1 ⇒h=1−−−−−−−−−(1)  \Rightarrow {h^2} = 1 \\\ \Rightarrow h = 1 - - - - - - - - - (1) \\\
Now we put (2,3)(2,3) in the equation of the circle, we get,
⇒(2−h)2+(3−k)2=k2\Rightarrow {(2 - h)^2} + {(3 - k)^2} = {k^2}
Now we solve the equation by applying the formula as (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab, we get
⇒4+h2−4h+9+k2−6k=k2\Rightarrow 4 + {h^2} - 4h + 9 + {k^2} - 6k = {k^2}
By solving we get
⇒h2+k2−4h−6k+13=k2\Rightarrow {h^2} + {k^2} - 4h - 6k + 13 = {k^2}
⇒h2−4h−6k+13=0−−−(2)\Rightarrow {h^2} - 4h - 6k + 13 = 0 - - - (2)
Now we put the value of h from equation (1) in equation (2), we get
⇒(1)2−4×1−6k+13=0 ⇒1−4+13=6k ⇒6k=10  \Rightarrow {(1)^2} - 4 \times 1 - 6k + 13 = 0 \\\ \Rightarrow 1 - 4 + 13 = 6k \\\ \Rightarrow 6k = 10 \\\
Form the equation we get the value of k
⇒k=106 ⇒k=53  \Rightarrow k = \dfrac{{10}}{6} \\\ \Rightarrow k = \dfrac{5}{3} \\\
So, the radius of the circle k=53k = \dfrac{5}{3}
We know that Diameter=2×radiusDiameter = 2 \times radius
⇒D=2×k\Rightarrow D = 2 \times k
Now we put the value of k in the formula we get
⇒D=2×53\Rightarrow D = 2 \times \dfrac{5}{3}
By solving we get the value of diameter D,
⇒D=103\Rightarrow D = \dfrac{{10}}{3}
So, the diameter of the circle is 103\dfrac{{10}}{3}.

So, the correct answer is “Option D”.

Note: The equation of the circle is the way of expressing definition of the circle in a coordinate plane. The equation is (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}. With the help of this equation we can find the radius of the circle and center as well.