Advertisements
Advertisements
Question
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
Advertisements
Solution
Given: Two interesting circles with centres C &D.
AB is their common chord.
To prove: AB bisected by CD at right angles.
Proof : CA = CB ...(radii)
∴ C lies on the right bisector of AB.
Similarly, D lies on the right bisector of AB.
Therefore, CD is the right bisector of AB.
Hence proved.
RELATED QUESTIONS
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of the moving end of the minute hand of a clock.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
- always 4 cm from the line AB.
- equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
ΔPBC and ΔQBC are two isosceles triangles on the same base BC but on the opposite sides of line BC. Show that PQ bisects BC at right angles.
Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?
