Equation of a Circle

Edexcel IAL WMA12/01/P2 /June 2021/Q6 (Equation of Circles)

A circle has equation​​ x2 – 6x + y2 + 8y + k = 0​​ 

where k is a positive constant.​​ 

Given that the x-axis is a tangent to this circle,​​ 

• find the value of k.​​ 

(3)

The circle meets the coordinate axes at the points R, S and T.

• Find the exact area of the triangle RST.​​ 

(4)

SOLUTION​​ 

a- Changing the equation to the standard circle equation.​​ 

x2-6x+y2+8y+k=0

Using the completung square method,​​ 

x-32-9+y+42-16+k=0

x-32+y+42-25+k=0

x-32+y+42=25-k

Comparing it to the standard circle equation.​​ 

Center 3, -4    r=25-k

Have a look at the sketch below.​​ 

Equating the radius with 4 since tangent to circle is x – axis, so the y-coordinate of centre point represents the radius.

25-k=4

25-k=16

25-16=k

k=9 

b-​​ At all 3 points R, S, & T, circle meets the axes. And, remember, x-axis is tangential to the circle. So, let R be the point where x-axis is a tangent to the circle. Therefore,​​ R 3,0. Whereas, the points where the y-axis meets the circle, must be having the x-cordinate 0. This would make circle equations to look like​​ 

x-32+y+42=16

0-32+y+42=16

9+y+42=7

y+4=±7

y=-4±7

Let us consider the x-coordinate of S and T be​​ -4+7​​ &​​ -4-7.

For understanding better the base abnd height of the triangle, it has been marked below.​​ 

The formula​​ 12×base×height​​ holds true for all triangles, not just right-angled. Once we have identified the base to be the difference between the two y-coordinates, the perpendicular height is simply the x-coordinate of the cut on the x-axis.

A=12 x ST x OR

For ST

ST= -4+7--4-7

ST=27

Whereas,​​ OR=3. Therefore, area comes out to be​​ 

A=12 x 27 x 3

A=37 sq. units