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Intersection of two circles

Intersection of two circles

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Explainer Video

Tutor: Bilal

Summary

Intersection of two circles

In a nutshell

To find the point at which two circles intersect, the equations for the two circles may be solved simultaneously. 


Example 1

The circle with equation (x−11)2+(y−2)2=100(x-11)^2+(y-2)^2=100(x−11)2+(y−2)2=100 and centre AAA intersects the circle with equation (x−7)2+(y−4)2=40(x-7)^2+(y-4)^2=40(x−7)2+(y−4)2=40 and centre BBB and points RRR and SSS.

Maths; Circles; KS5 Year 12; Intersection of two circles

a) Find the intersection points RR R​ and SSS.


First, expand the equation for the circle with centre AAA, and rearrange to make the equation equal to 000.

(x−11)2+(y−2)2=100x2−22x+121+y2−4y+4=100x2−22x+y2−4y+25=0\begin{aligned}(x-11)^2 + (y-2)^2 &= 100\\x^2-22x+121+y^2-4y+4&=100\\x^2-22x+y^2-4y+25&=0 \end{aligned}(x−11)2+(y−2)2x2−22x+121+y2−4y+4x2−22x+y2−4y+25​=100=100=0​


Do the same for circle with centre BBB.

(x−7)2+(y−4)2=40x2−14x+49+y2−8y+16=40x2−14x+y2−8y+25=0\begin{aligned}(x-7)^2+(y-4)^2&=40 \\x^2-14x+49+y^2-8y+16&=40\\x^2-14x+y^2-8y+25&=0 \end{aligned}(x−7)2+(y−4)2x2−14x+49+y2−8y+16x2−14x+y2−8y+25​=40=40=0​​​


Solve the equations and make yyy the subject. ​

x2−22x+y2−4y+25=x2−14x+y2−8y+25−22x−4y+25=−14x−8y+254y=8xy=2x\begin{aligned}x^2-22x+y^2-4y+25&=x^2-14x+y^2-8y+25 \\-22x-4y+25&=-14x-8y+25\\4y&=8x \\y&=2x \end{aligned}x2−22x+y2−4y+25−22x−4y+254yy​=x2−14x+y2−8y+25=−14x−8y+25=8x=2x​​​


Now, substitute the above equation into x2−22x+y2−4y+25=0x^2-22x+y^2-4y+25=0x2−22x+y2−4y+25=0.

x2−22x+(2x)2−4(2x)+25=0x2−22x+4x2−8x+25=05x2−30x+25=0x=−−30±(−30)2−(4×5×25)2×5x=30±900−50010x=30±40010x=5,1\begin{aligned}x^2-22x+(2x)^2-4(2x)+25&=0 \\x^2-22x+4x^2-8x+25&=0 \\5x^2-30x+2&5=0 \\x&=\dfrac{--30\pm\sqrt{(-30)^2-(4\times5\times25)}}{2\times5} \\\\x&= \dfrac{30\pm\sqrt{900-500}}{10} \\\\x&= \dfrac{30\pm\sqrt{400}}{10} \\\\x&=5, 1 \end{aligned}x2−22x+(2x)2−4(2x)+25x2−22x+4x2−8x+255x2−30x+2xxxx​=0=05=0=2×5−−30±(−30)2−(4×5×25)​​=1030±900−500​​=1030±400​​=5,1​​​


Finally, substitute values for xxx​ into y=2xy=2xy=2x​ to solve for yyy to find points RRR and SSS.

y=2(5)y=10y=2(1)y=2\begin{aligned}y&=2(5)\\y&=10\\\\y&=2(1)\\y&=2\end{aligned}yyyy​=2(5)=10=2(1)=2​

Therefore, points RRR and SSS are (5,10)‾\underline{(5,10)}(5,10)​ and (1,2)‾\underline{(1,2)}(1,2)​, respectively.​


b) Find the area of the right-angled triangle RBSRBSRBS.

Maths; Circles; KS5 Year 12; Intersection of two circles

To find the area of the triangle, the lengths of the sides BR,BSBR, BSBR,BS and RSRSRS must be found. As points RRR and SSS lie on the circle with centre BBB.

BR=BS=r=40=210BR=BS=r = \sqrt{40} = 2\sqrt{10}BR=BS=r=40​=210​​​


To find the length of RSRSRS, use d=(x2−x1)2+(y2−y1)2d=\sqrt{{(x_{2} - x_{1}})^2+{(y_{2} - y_{1})^2}}d=(x2​−x1​)2+(y2​−y1​)2​ for points R(5,10)R(5,10)R(5,10) and S(1,2)S(1,2)S(1,2).

  d=(x2−x1)2+(y2−y1)2=(1−5)2+(2−10)2=80=45\begin{aligned}d&=\sqrt{{(x_{2} - x_{1}})^2+{(y_{2} - y_{1})^2}} \\&=\sqrt{{(1 - 5})^2+{(2 - 10)^2}}\\&=\sqrt{80} \\&=4\sqrt5\end{aligned}d​=(x2​−x1​)2+(y2​−y1​)2​=(1−5)2+(2−10)2​=80​=45​​


Use the cosine rule to find the angle ∠RBS\angle RBS∠RBS:

cos⁡x=b2+c2−a22bccos⁡x=(210)2+(210)2−(45)22×210×210cos⁡x=0x=cos⁡−1(0)x=90°\begin{aligned}\cos x &= \dfrac {b^2+c^2-a^2}{2bc} \\\\\cos x&= \dfrac{(2\sqrt{10})^2 + (2\sqrt{10})^2 - (4\sqrt{5})^2}{2 \times2\sqrt{10} \times 2\sqrt{10} } \\\\ \cos x &=0 \\\\x &= \cos^{-1} (0) \\x&= 90 \degree\end{aligned}cosxcosxcosxxx​=2bcb2+c2−a2​=2×210​×210​(210​)2+(210​)2−(45​)2​=0=cos−1(0)=90°​​​


As triangle RBSRBSRBS is right angled, find the area using A=12bhA = \dfrac{1}{2} bhA=21​bh:​

​​​A=12bh=12×210×210=20\begin{aligned}A &= \dfrac{1}{2}bh \\\\&= \dfrac{1}{2} \times 2 \sqrt{10} \times 2 \sqrt{10} \\&=20 \end{aligned}A​=21​bh=21​×210​×210​=20​​​

​​


Therefore, the area of the triangle is 20‾\underline{20 }20​.


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Intersection of straight lines and circles

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FAQs - Frequently Asked Questions

What is the centre of a circle with the equation (x-a)^2+(y-b)^2=r^2?

The centre of a circle with the equation (x-a)^2+(y-b)^2=r^2 is (a,b).

How can the distance between two points be found?

To find the distance between two points (x1,y1) and (x2,y2), find the square root of ((x1 - x2)^2+(y1 - y2)^2).

How can the intersection points of two circles be found?

To find the points at which two circles intersect, the equations for the two circles may be solved simultaneously. 

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