Let C be a circle passing through the points A(2, –1) and B (3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle
\((x−5)^2+(y−1)^2=\frac{13}{2}\)
then r2 is equal to
Let C be a circle passing through the points A(2, –1) and B (3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle
\((x−5)^2+(y−1)^2=\frac{13}{2}\)
then r2 is equal to
32
\(\frac{65}{2}\)
\(\frac{61}{2}\)
30
The Correct Option is B
Solution and Explanation
The correct answer is (B) : \(\frac{65}{2}\)
Equation of perpendicular bisector of AB is
\(y−32=−\frac{1}{5}(x−52)⇒x+5y=10\)
Solving it with equation of given circle,
\((x−5)^2+(\frac{10−x}{5}−1)^2=\frac{13}{2}\)
\(⇒(x−5)^2(1+\frac{1}{25})=\frac{13}{2}\)
\(⇒x−5=±\frac{5}{2}⇒x=\frac{5}{2} or \frac{15}{2}\)
But \(x≠\frac{5}{2}\)
because AB is not the diameter.
So, centre will be \((\frac{15}{2},\frac{1}{2})\)
Now \(r^2=(\frac{15}{2}−2)^2+(\frac{1}{2}+1)^2\)
\(=\frac{65}{2}\)
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Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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