3. Draw a circle with centre 0 and any convenient radius. Let this circle be C,. Take a point
A on circle C, and join OA. Now take any point B on OA. With centre A and radius = AB
draw a circle and call it Cz. Let C and D be the points of intersection of the two circles,
c, and Cz
Name:
i. Three points on circle C
ii. Three points on circle C,
lii. A point in the interior of circle C, and on circle Cz.
iv. A point lying in the exterior of circle Cz.
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Answers
Answer:
A on circle C, then sign up for OA. Any point B on OA right now. A centre A and a radius equal to AB.
Step-by-step explanation:
Step : 1 The circle with a radius of 5 cm and chord A is what we are requested to design.
We do the subsequent actions:
STEP 1: Sketch a circle with a radius of 5 cm and a centre at point O.
Step 2 is to pull the cord AB.
STEP 3: Draw two arcs, one on each side of AB, with the centre at A and the radius being greater than half of AB.
STEP 4: Draw arcs at C and D that intersect the arcs generated in step 3 with B as the centre and the same radius as in step 3.
Step : 2 Draw the line segment using C and D as the endpoints in step 5.
The image up above displays the diagram.
The necessary perpendicular bisector of AB is the line segment CD. Since CD is the perpendicular bisector of AB, which is the circle's chord, the circle's centre is where it intersects.
Step : 3 Consider the equation (xxc)2+(yyc)2r2=0 for a circle. .
Three quadratic equations are formed by plugging in the three points: (1xc)2+(1yc)2r2=0.
Create two linear equations by subtracting the first from the second and the first from the third.
As (xc,yc)=, solve for the centre (3,2)
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