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Question: If one of the diameters of the circle \({x^2} + {y^2} - 2x - 6y + 6 = 0\) is a chord to the circle w...

If one of the diameters of the circle x2+y2−2x−6y+6=0{x^2} + {y^2} - 2x - 6y + 6 = 0 is a chord to the circle with centre (2,1), the radius of the circle is
A. 3\sqrt 3 .
B.2\sqrt 2 .
C. 3.
D. 2.

Explanation

Solution

To solve this question, we will use some basic concepts of circles. The equation of circle with centre (h,k) and radius r is given by, (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} and we will also use the distance formula that is given as, d=(x2−x1)2+(y2−y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} .

Complete step-by-step answer:
Given that,
Diameter of circle x2+y2−2x−6y+6=0{x^2} + {y^2} - 2x - 6y + 6 = 0 is a chord to the circle with centre (2,1).
⇒x2+y2−2x−6y+6=0 ⇒x2−2x+y2−6y=−6 ⇒x2−2x+(1)2+y2−6y+(3)2=−6+(1)2+(3)2=−6+1+9 ⇒(x−1)2+(y−3)2=4  \Rightarrow {x^2} + {y^2} - 2x - 6y + 6 = 0 \\\ \Rightarrow {x^2} - 2x + {y^2} - 6y = - 6 \\\ \Rightarrow {x^2} - 2x + {\left( 1 \right)^2} + {y^2} - 6y + {\left( 3 \right)^2} = - 6 + {\left( 1 \right)^2} + {\left( 3 \right)^2} = - 6 + 1 + 9 \\\ \Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - 3} \right)^2} = 4 \\\
⇒(x−1)2+(y−3)2=(2)2\Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - 3} \right)^2} = {\left( 2 \right)^2}. ……….. (i)
As we know that equation of circle is given by,
(x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}.
Comparing equation (i) with this, we will get
Center of circle (h,k) as (1,3).
Now all these information can be represented diagrammatically as,


Here, we can see that AB is the diameter and D is the centre of the circle x2+y2−2x−6y+6=0{x^2} + {y^2} - 2x - 6y + 6 = 0.
Now, the radius of the circle x2+y2−2x−6y+6=0{x^2} + {y^2} - 2x - 6y + 6 = 0 , DB is 2 units.
Now, we have to find out the length of OD.
We know that, the distance between two points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is given by,
d=(x2−x1)2+(y2−y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} .
We have, O = (2,1) and D = (1,3).
So, the length of OD will be,
⇒OD=(2−1)2+(1−3)2 ⇒OD=1+4 ⇒OD=5  \Rightarrow OD = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 3} \right)}^2}} \\\ \Rightarrow OD = \sqrt {1 + 4} \\\ \Rightarrow OD = \sqrt 5 \\\
Now, we have OD=5OD = \sqrt 5 units and DB = 2 units.
Hence, by using Pythagoras theorem, we will find out OB.
⇒OB=(OD)2+(DB)2 ⇒OB=(5)2+(2)2 ⇒OB=5+4=9 ⇒OB=3  \Rightarrow OB = \sqrt {{{\left( {OD} \right)}^2} + {{\left( {DB} \right)}^2}} \\\ \Rightarrow OB = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( 2 \right)}^2}} \\\ \Rightarrow OB = \sqrt {5 + 4} = \sqrt 9 \\\ \Rightarrow OB = 3 \\\
From the figure, we can clearly see that OB is the required radius of the circle.
Hence, the correct answer is option (C).

Note: Whenever we ask such questions, we have to remember the method of the equation of the circle. First, we have to find out the radius and centre of the circle from the given equation of circle. Then by using the distance formula, we will get the length of the chord. After that by making the diagram from the given details, we can easily identify the required radius of the circle and using the Pythagoras theorem, we will get the required answer.