Working will be shown later

i)

Touches the axes means they are tangent to both the x and y-axis. Touches only once for each.

Since we know y = x passes through the centres of C1 and C2 ,

Their centres have equal x and y coordinates.

(In fact, touches the axes already implies this)

So use the equation of the circle :

(x - a)² + (y - b)² = r², where (a,b) are the coordinates of the centre and r is the radius.

Since the x and y coordinates are equal, a = b

Since they are tangent to the axes, their centres are equidistant from either axis.

Why?

Because, from circle properties we know that tangent is perpendicular to radius.

(Also, tangents from an external point are equal in length. In this case, the external point is the origin (0,0))

Since the x and y-axis are perpendicular to each other,

we know that each centre is 1 radius away (5 units) from either axis.

(5 units horizontally from y-axis, 5 units vertically from x-axis)


For each circle, we can visualise a square formed by the joining the origin, the two tangent points and the centre.


In essence , this means that the x and y coordinates are either (5,5) or (-5,5)

So, sub a = b = 5 (and a = b = -5) , r = 5,

we have the two equations for C1 and C2 :


(x - 5)² + (y - 5)² = 5² →(x - 5)² + (y - 5)² = 25


(x - (-5))² + (y - (-5))² = 5² →(x + 5)² + (y + 5)² = 25


Question did not mention which is which so no need to label C1 or C2.

ii)

It is given that the centre of C3 is 6 units away from the line joining the centres of C1 and C2.

We take this to refer to the perpendicular distance from the line y = x to the centre of C3.

(Part of the line y = x joins two centres of C1 and C2, since they both lie on it)


This means that C3 lies on a line perpendicular to y = x. This line has equation y = -x + c, where it appears that c is unknown and to be found.


But recall the property that a perpendicular bisector of a chord passes through the centre.

The portion of line y = x joining C1's centre to C2's basically a chord of C3.


And since the two centres are equidistant from the origin, the origin is the midpoint of the chord and this tells us that the perpendicular bisector passes through it.


(Perpendicular bisector means it cuts the chord exactly into two equal parts and is at a right angle to it. So it must pass through all points equidistant from either end of the chord, including the chord's midpoint)

So , the y-intercept, c = 0. The perpendicular bisector has equation y = -x

And since it passes through the centre of C3 as well , the centre lies on y = -x and this tells us that its coordinates are negatives of each other.

Since the centre's x and y-coordinates are negatives of each other and lie on y = -x, it is also perpendicularly equidistant from either axes.


Draw a horizontal line from the y-axis and vertical line from x-axis to the centre of C3.

We get a square formed, where the centre, the origin , and those two intersection points with the axes are its vertices.


Essentially there are two right-angled triangles and we can apply Pythagoras' Theorem on either triangle

(or linear distance formula, which is essentially based on this theorem)


The hypotenuse is 6 units long since its length is just the perpendicular distance from the centre to the origin.


Let (A,-A) be the coordinates of C3's centre.

Vertical and horizontal distance from either axis
= |A| units

(because A could be negative or positive, we put the modulus)


Using Pythagoras' Theorem,

|A|² +|A|² = 6²
2|A|² = 36
2A² = 36
A² = 18
A = ±√18 = ±3√2


So the centre could either be (3√2, -3√2) or (-3√2, 3√2)



If you use linear distance formula it will be like this :


√((A - 0)² + (-A - 0)²) = 6
√(A² + (-A)²) = 6
√(2A²) = 6
√A² = 6/√2 = (3×2)/√2 = 3√2

|A| = 3√2

A = 3√2 or -3√2

Lastly , find the radius of C3.

Sub (5,5) as your (x,y) and (3√2, -3√2) as centre coordinates (a,b)

(You can use (-5,-5) and (-3√2,3√2) also. Any combination of a pair will do)



(5 - 3√2)² + (5 - (-3√2))² = r²

(5 - 3√2)² + (5 + 3√2)² = r²


This is in the form (a - b)² + (a + b)² which equals 2a² + 2b²

(shortcut, so no need to expand the expressions)


2(5²) + 2(3√2)² = r²
50 + 36 = r²
r² = 86


Write the equations accordingly.



Alternatively, we notice on the diagram that a rhombus is formed by joining the origin, the centres of C1 and C2 and both possible C3's.


There are 4 right angled triangles inside.

We can use Pythagoras' theorem as well


r² = (distance from origin to (5,5))² + (distance from origin to (3√2, -3√2))²

r² = (5² + 5²) + ((3√2)² + (-3√2)²)

r² = 25 + 25 + 18 + 18 = 86


Pythagoras' Theorem was applied twice here.