Question

A circle with centre at the point (2,-1) passes through the point A at (4,-5).

(a) Find an equation for the circle C.

(b) Find an equation of the tangent to the circle C at the point A, giving your answer in the form ax+by+c=0, where a, b, and c are integers..


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Answer 1

Explanation:

(b) Find an equation of the tangent to the circle C at the point A, giving your answer in the form ax+by+c=0, where a, b, and c are integers..

Answer 2

a)

`(x - 2) {}^(2) + (x +1) {}^(2) = 20`

b)

`x - 2y - 14 = 0`

Explanation:

a) the centre is (2,-1)

the equation so far is:

`(x - 2) {}^(2) + (x +1) {}^(2) = r {}^(2)`

because you just change the sign of 2 to -2 and change the sign of -1 to +1

to find the radius:

`r = \sqrt{(change \: in \: x) {}^(2) + (change \: in \: y) {}^(2) }`

`r = \sqrt{(4 - 2) {}^(2) + ( - 5 - ( - 1)) {}^(2) }`

`the \: radius \: is \: 2 √(5)`

to find diameter square 2√5

`(2 √(5) \:) {}^(2) = 20 \\ the \: equation \: is(x - 2) {}^(2) + (x +1) {}^(2) = 20`

b) find the gradient between the points (2,-1) and (4,-5)

`gradient = (change \: in \: y)/(change \: in \: x)`

`\: gradient = ( - 5 + 1)/(4 - ( - 1)) = - 2`

the gradient of the tangent is the negative reciprocal of -2. This means you flip it upside down which would be -1/2, then multiply by -1 which is 1/2

the gradient is 1/2

using the gradient and the point (4,-5) to write the equation in the form y=mx+c

here y is -5, m is the gradient (1/2) and x is 4

we need to work out c

`y = mx + c`

`- 5 = (1)/(2) (4) + c`

`- 5 = 2 + c`

subtract 2 on both sides

`- 5 - 2 = c`

`c = - 7`

now the equation is

`y = (1)/(2) x - 7`

but we need to write it in the form ax+by+c=0

subtract y on both sides:

`(1)/(2) x - y - 7 = 0`

multiply everything by 2 to get rid of the fraction

`x - 2y - 14 = 0`

that is the final equation