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Question: Find the equation of a circle whose center is \(\left( { - 3,1} \right)\) and which passes through t...

Find the equation of a circle whose center is (−3,1)\left( { - 3,1} \right) and which passes through the point (5,2)\left( {5,2} \right).

Explanation

Solution

The given problem tests us on the concepts of coordinate geometry as well as conic sections. The problem requires us to find the equation of a circle given certain conditions. We first have to analyze the problem by drawing a figure. We are given the centre of the circle and a point lying on it. So, we will have to calculate the radius of the circle first by using the distance formula and then substitute the values of centre and radius of the circle in the standard equation of the circle.

Complete step-by-step solution:
We are given the center of the circle as (−3,1)\left( { - 3,1} \right).
Also, the circle passes through the point (5,2)\left( {5,2} \right) .
Now, we know that the radius of a circle is the line segment joining any point on the circumference of the circle to the center of the circle.

We know that we can find the distance between any two points using the distance formula (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} where (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) are the coordinates of the two points.
Now, we find the length of radius of the circle by calculating the distance between the centre of the circle and the point lying on the circumference of the circle.
So, radius of the circle=(5−(−3))2+(2−1)2= \sqrt {{{\left( {5 - \left( { - 3} \right)} \right)}^2} + {{\left( {2 - 1} \right)}^2}}
=(8)2+(1)2= \sqrt {{{\left( 8 \right)}^2} + {{\left( 1 \right)}^2}} units
=64+1= \sqrt {64 + 1} units
=65= \sqrt {65} units
So, the length of the radius of the circle is 65\sqrt {65} units
Now, we know that the standard equation of a circle whose centre is known to us as (α,β)\left( {\alpha ,\beta } \right) and the radius is r units is (x−α)2+(y−β)2=r2{\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}.
So, we know that the centre of the circle is (−3,1)\left( { - 3,1} \right) and the radius is 65\sqrt {65} units.
So, we get the equation of the circle as (x−(−3))2+(y−1)2=(65)2{\left( {x - \left( { - 3} \right)} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {\sqrt {65} } \right)^2}
Opening the brackets and simplifying the equation, we get,
⇒(x+3)2+(y−1)2=65\Rightarrow {\left( {x + 3} \right)^2} + {\left( {y - 1} \right)^2} = 65
So, the equation of the required circle whose center is (−3,1)\left( { - 3,1} \right) and which passes through the point (5,2)\left( {5,2} \right) is (x+3)2+(y−1)2=65{\left( {x + 3} \right)^2} + {\left( {y - 1} \right)^2} = 65.

Note: The given question also involves basic understanding of equations of conic sections. Such problems illustrate the interdependence of mathematical ideas and topics on each other. We know that the diameter of a circle is twice the length of radius of a circle. To find the equation of the required circle, we should know the radius and centre of the circle.