
Radius of circle ( x – 5 ) ( x – 1 ) + ( y – 7 ) ( y – 4 ) = 0 is
(A) 3
(B) 4
(C) $\dfrac{5}{2}$
(D) $\dfrac{7}{2}$
Answer
162k+ views
Hint: In this question, we have to find the radius of the circle. First we solve the given equation then we know the standard form of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. By using this formula and comparing the values, we get the value of g, f and c. Then we put these values in the formula of radius and get our desired answer.
Complete Step by step solution:
Given equation is (x-5) (x-1) + (y-7) (y-4) =0 -------------------- (1)
We have to find the radius of the circle.
By expanding the equation (1), we get
${{x}^{2}}-6x+5+{{y}^{2}}-11y+28=0$
That is ${{x}^{2}}+{{y}^{2}}-6x-11y+33=0$----------------------------- (2)
We know the standard form of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
Npw compare the equation (2) with the standard form of circle, we get
2gx = - 6x and 2fy = - 11 y
We get g = -3 and f = $\dfrac{5}{2}$$\dfrac{-11}{2}$ and c = 33
From there, we get the centre = ( -g,f )
We know radius = $\sqrt{{{g}^{2}}+{{f}^{2}}-2gf}$
Now put the values of g and f in the above formula, we get
R = $\sqrt{{{(-3)}^{2}}+{{\left( \dfrac{-11}{2} \right)}^{2}}+2(-3)\left( \dfrac{-11}{2} \right)}$
= $\sqrt{9+\dfrac{121}{4}-33}$
= $\sqrt{\dfrac{25}{4}}$ = $\dfrac{5}{2}$
From there, we get radius of circle = $\dfrac{5}{2}$
Thus, Option ( C ) is correct.
Note: In these types of questions, Students made mistakes in comparing the values. If we know the proper standard form, then we are able to solve these types of questions. We should remember the formula properly to get the correct answer. With a lot of practice, students must be able to solve the problem correctly.
Complete Step by step solution:
Given equation is (x-5) (x-1) + (y-7) (y-4) =0 -------------------- (1)
We have to find the radius of the circle.
By expanding the equation (1), we get
${{x}^{2}}-6x+5+{{y}^{2}}-11y+28=0$
That is ${{x}^{2}}+{{y}^{2}}-6x-11y+33=0$----------------------------- (2)
We know the standard form of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
Npw compare the equation (2) with the standard form of circle, we get
2gx = - 6x and 2fy = - 11 y
We get g = -3 and f = $\dfrac{5}{2}$$\dfrac{-11}{2}$ and c = 33
From there, we get the centre = ( -g,f )
We know radius = $\sqrt{{{g}^{2}}+{{f}^{2}}-2gf}$
Now put the values of g and f in the above formula, we get
R = $\sqrt{{{(-3)}^{2}}+{{\left( \dfrac{-11}{2} \right)}^{2}}+2(-3)\left( \dfrac{-11}{2} \right)}$
= $\sqrt{9+\dfrac{121}{4}-33}$
= $\sqrt{\dfrac{25}{4}}$ = $\dfrac{5}{2}$
From there, we get radius of circle = $\dfrac{5}{2}$
Thus, Option ( C ) is correct.
Note: In these types of questions, Students made mistakes in comparing the values. If we know the proper standard form, then we are able to solve these types of questions. We should remember the formula properly to get the correct answer. With a lot of practice, students must be able to solve the problem correctly.
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