There is a circle on the co -ordinate plane with two tangent lines 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 with two tangent lines 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 (3/4)√10,(3/4, -3/4)
Let us first look at the nature of the 2 lines. We observe that the coefficients of x and y for both the lines in 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 to 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 a point of contact. Tangents PQ and XY are perpendicular to radii OA and OB respectively. So ∠PAO and ∠ XBO are of 90∘.
2 lines that are perpendicular to 2 parallel lines are at 180∘ to each other. Line segments OA and OB are perpendicular to parallel lines PQ and XY. So, OA and OB are at 180∘. Thus, AOB is at180∘ and AOB is a straight line segment.
AOB is a line segment passing centre O. Also A and B are points on the circumference. Thus, AB is the diameter of a 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 the centre. The distance AB is the diameter. So half of distance AB is the radius.
A = (32,32)
B= (0,-3)
Centre is midpoint of AB = (34,−34)
Applying distance formula
AB= [(x2−x1)2 + (y2−y1)2]
AB = [(32−0)2 + (32+3)2 ]
AB = 32√10
Radius is half of AB = 34√10
The midpoint of 2 points in coordinate geometry is
(3/4)√10,(3/4, -3/4)
Let us first look at the nature of the 2 lines. We observe that the coefficients of x and y for both the lines in 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 to 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 a point of contact. Tangents PQ and XY are perpendicular to radii OA and OB respectively. So ∠PAO and ∠ XBO are of 90∘.
2 lines that are perpendicular to 2 parallel lines are at 180∘ to each other. Line segments OA and OB are perpendicular to parallel lines PQ and XY. So, OA and OB are at 180∘. Thus, AOB is at180∘ and AOB is a straight line segment.
AOB is a line segment passing centre O. Also A and B are points on the circumference. Thus, AB is the diameter of a 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 the centre. The distance AB is the diameter. So half of distance AB is the radius.
A = (32,32)
B= (0,-3)
Centre is midpoint of AB = (34,−34)
Applying distance formula
AB= [(x2−x1)2 + (y2−y1)2]
AB = [(32−0)2 + (32+3)2 ]
AB = 32√10
Radius is half of AB = 34√10
The midpoint of 2 points in coordinate geometry is
