There is a circle on the co- ordinate plane for which there are 2 tangent lines. These are x - y = 3 and x - y = 0. The meeting points of the circle with the 2 lines are (32,32) and (0,3). Find the radius and the centre of the circle.
There is a circle on the co- ordinate plane for which there are 2 tangent lines. These are x - y = 3 and x - y = 0. The meeting points of the circle with the 2 lines are (32,32) and (0,3). Find the radius and the centre of the circle.
The correct option is A 4.5√2 , (3/4, 9/4)
Let us first look at the nature of the 2 lines. We observe that the coefficients of x and y for both the lines� equations are same. But the constant term is different for both terms.
Let us use the concepts in linear equations in 2 variables.
Let us consider 2 lines with equations given below.
The above condition is satisfied by the 2 lines given in the question. This means that the 2 lines are parallel.
In other words we have 2 parallel tangents for a circle.
This can be represented by the diagram below.
Let PQ and XY be the 2 tangents. A and B are the 2 points of contact with the circle respectively. The centre of the circle is O. Join OA and OB.
Tangents are perpendicular to radii at point of contact. Tangents PQ and XY are perpendicular to radii OA and OB respectively. So angles PAO and angles XBO are of 90 degrees.
2 lines that are perpendicular to 2 parallel lines are at 180 degrees to each other. Line segments OA and OB are perpendicular to parallel lines PQ and XY. So, OA and OB are at 180 degrees. Thus AOB is at 180 degrees and AOB is a straight line segment.
AOB is a line segment passing centre O. Also A and B are points on circumference. Thus, AB is diameter of circle and its midpoint is centre O.
We already know the coordinates of contact points A and B. The midpoint of A and B will give us centre. The distance AB is diameter. So half of distance AB is radius.
A = (32,32)
B= (0,3)
Applying distance formula-
AB= ^a [(x2−x1)2 + (y2−y1)2]
AB = ^a [ (32−0)2 + (32−3)2 ]
AB = 4.5 a2
The midpoint of 2 points in coordinate geometry is
4.5√2 , (3/4, 9/4)
Let us first look at the nature of the 2 lines. We observe that the coefficients of x and y for both the lines� equations are same. But the constant term is different for both terms.
Let us use the concepts in linear equations in 2 variables.
Let us consider 2 lines with equations given below.
The above condition is satisfied by the 2 lines given in the question. This means that the 2 lines are parallel.
In other words we have 2 parallel tangents for a circle.
This can be represented by the diagram below.
Let PQ and XY be the 2 tangents. A and B are the 2 points of contact with the circle respectively. The centre of the circle is O. Join OA and OB.
Tangents are perpendicular to radii at point of contact. Tangents PQ and XY are perpendicular to radii OA and OB respectively. So angles PAO and angles XBO are of 90 degrees.
2 lines that are perpendicular to 2 parallel lines are at 180 degrees to each other. Line segments OA and OB are perpendicular to parallel lines PQ and XY. So, OA and OB are at 180 degrees. Thus AOB is at 180 degrees and AOB is a straight line segment.
AOB is a line segment passing centre O. Also A and B are points on circumference. Thus, AB is diameter of circle and its midpoint is centre O.
We already know the coordinates of contact points A and B. The midpoint of A and B will give us centre. The distance AB is diameter. So half of distance AB is radius.
A = (32,32)
B= (0,3)
Applying distance formula-
AB= ^a [(x2−x1)2 + (y2−y1)2]
AB = ^a [ (32−0)2 + (32−3)2 ]
AB = 4.5 a2
The midpoint of 2 points in coordinate geometry is
