Question

one circle has radius 5 and its centre (0,5). A second circle has radius of 12 and its centre (12,0). What is the length of the radius of a third circle which passes through the centre of the second circle and both the points of intersection of the first two circles.

Answer 1

The point made triangle... And the angle at origin is 90..so radious would be R2=144+25=169
Therefore R=13

Answer 2

Answer:

6.5 cm

Step-by-step explanation:

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122 x2 + y2 - 24x = 0

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122 x2 + y2 - 24x = 0Now third circle is passing through these two circles intersection points so it will be

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122 x2 + y2 - 24x = 0Now third circle is passing through these two circles intersection points so it will be x2 + y2 - 10y + λ(x2 + y 2 - 24x) = 0

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122 x2 + y2 - 24x = 0Now third circle is passing through these two circles intersection points so it will be x2 + y2 - 10y + λ(x2 + y 2 - 24x) = 0Also third circle is passing through point (12,0) so putting this point in above equationwe get λ = 1 and put this value we get equation of third circle,

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122 x2 + y2 - 24x = 0Now third circle is passing through these two circles intersection points so it will be x2 + y2 - 10y + λ(x2 + y 2 - 24x) = 0Also third circle is passing through point (12,0) so putting this point in above equationwe get λ = 1 and put this value we get equation of third circle,x2 + y2 - 12x + 5y = 0.

For first circle centre is (0 , 5) and radius is 5 so equation of the circle,(x - 0)2 + (y - 5)2 = 52 x2 + y2 - 10y = 0For second circle centre is (12, 0) and radius is 12 so,(x - 12)2 + (y - 0)2 = 122 x2 + y2 - 24x = 0Now third circle is passing through these two circles intersection points so it will be x2 + y2 - 10y + λ(x2 + y 2 - 24x) = 0Also third circle is passing through point (12,0) so putting this point in above equationwe get λ = 1 and put this value we get equation of third circle,x2 + y2 - 12x + 5y = 0.This gives centre (6 , 5/2) and radius = 6.5

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