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Circles - Circle through Three Points

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Uniqueness of a Circle: There is one and only one circle passing through three given non-collinear points. If you imagine three points AA, BB, and CC that do not lie on a single straight line, they form the vertices of a triangle, and exactly one circle can be drawn to touch all three vertices.

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Non-existence for Collinear Points: No circle can pass through three points if they are collinear (lying on the same straight line). Visually, a circle has a constant curvature, whereas a straight line has no curvature; therefore, a circle cannot satisfy the linear alignment of three collinear points.

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Perpendicular Bisector Method: To locate the center of a circle passing through three points PP, QQ, and RR, you must construct the perpendicular bisectors of any two segments, such as PQPQ and QRQR. The point where these two bisector lines intersect is the unique center OO of the circle.

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The Circumcenter: The center OO of the circle passing through the three vertices of a triangle is called the circumcenter. Visually, this point is equidistant from all three vertices, meaning OA=OB=OCOA = OB = OC.

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Infinite Circles through Two Points: Through two distinct points AA and BB, an infinite number of circles can be drawn. The centers of all these circles lie on the perpendicular bisector of the line segment ABAB. It is only the addition of a third non-collinear point that fixes the circle's size and position.

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Circumcircle and Circumradius: The circle passing through the vertices of a triangle is known as the circumcircle of the triangle, and its radius is referred to as the circumradius. In a right-angled triangle, the circumcenter lies exactly at the midpoint of the hypotenuse.

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Geometric Construction: When drawing a circle through points AA, BB, and CC, the perpendicular bisector of chord ABAB and chord BCBC will always meet at the center OO. This is because the center of a circle must lie on the perpendicular bisector of any chord of that circle.

📐Formulae

Distance from center OO to points: OA=OB=OC=rOA = OB = OC = r

Perpendicular Bisector Property: If OM⊥ABOM \perp AB and MM is the midpoint, then AM=BM=12ABAM = BM = \frac{1}{2}AB

General Circle Equation: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center OO

Condition for Collinearity: Three points AA, BB, and CC are collinear if AB+BC=ACAB + BC = AC (no circle can pass through them)

💡Examples

Problem 1:

Given three non-collinear points AA, BB, and CC. Describe the geometric procedure to find the center of the circle passing through these points.

Solution:

Step 1: Join the points to form line segments ABAB and BCBC. These segments act as chords of the required circle. \ Step 2: Construct the perpendicular bisector of segment ABAB. Any point on this line is equidistant from AA and BB. \ Step 3: Construct the perpendicular bisector of segment BCBC. Any point on this line is equidistant from BB and CC. \ Step 4: Find the point of intersection of these two perpendicular bisectors and label it OO. \ Step 5: Since OO lies on both bisectors, OA=OBOA = OB and OB=OCOB = OC. Therefore, OA=OB=OC=rOA = OB = OC = r. \ Step 6: With OO as the center and OAOA as the radius, draw the circle.

Explanation:

The solution relies on the theorem that the perpendicular bisector of a chord passes through the center of the circle. The intersection of two such bisectors uniquely identifies the center point equidistant from all three vertices.

Problem 2:

If the distance between the center O(0,0)O(0,0) and a point A(3,4)A(3,4) on the circle is rr, what is the radius of the circle passing through AA, and will it pass through B(−5,0)B(-5,0) and C(0,5)C(0,5)?

Solution:

Step 1: Calculate the radius rr using the distance formula for OAOA: r=(3−0)2+(4−0)2=32+42=9+16=25=5r = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ Step 2: Check if point B(−5,0)B(-5, 0) lies on the circle by calculating OBOB: OB=(−5−0)2+(0−0)2=(−5)2=5OB = \sqrt{(-5-0)^2 + (0-0)^2} = \sqrt{(-5)^2} = 5 \ Step 3: Check if point C(0,5)C(0, 5) lies on the circle by calculating OCOC: OC=(0−0)2+(5−0)2=52=5OC = \sqrt{(0-0)^2 + (5-0)^2} = \sqrt{5^2} = 5 \ Step 4: Since OA=OB=OC=5OA = OB = OC = 5, the circle with center (0,0)(0,0) and radius 55 passes through all three points.

Explanation:

To determine if a circle passes through three points, we verify if all three points are at the same distance (the radius) from a common center point. Since all three distances equal 55, they lie on the same unique circle.