Question

A circle has radius equal to 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through (7, 3).❤​

Answer 1

Answer:

The two possible equations for the circle will be $x^{2} +y^{2} -14x-12y+76=0$ and $x^{2} +y^{2} -8x-6y+16=0$

Explanation:

It is stated that the center of the circle lies on the line $y=x-1$
So, the center of circle must satisfy the equation of the line.

Let 'x' co-ordinate of the center to be 'h'
The 'y' co-ordinate will be 'h-1'

The equation of a circle is in the form - $(x-x_{1})^{2}+(y-y_{1})^{2}=r^{2}$
Here, $x_{1}$ and $y_{1}$ denote the center and 'r' denotes the radius.

The equation of the circle becomes - $[x-h]^{2}+[y-(h-1)]^{2}=3^{2}$

(Insert the values of 'x' and 'y' as 7 and 3 as the circle passes through that point)

Simplify it to get a quadratic equation with the variable 'h'.

$h^{2}-11h+28=0$

Solve for 'h' :-

$h = \dfrac{11\pm\sqrt{121-4(28)}}{2}=\dfrac{11\pm3}{2}=7,4$

Substitute the values of 'h' in the equation of the circle to get the equation for the circle:-

$1. \ (x-7)^{2} +(y-6)^{2}=9 \\x^{2} +y^{2}-14x-12y+76=0 \\ \\2. \ (x-4)^{2}+(y-3)^{2}=9 \\x^{2}+y^{2}-8x-6y+16=0$

Answer 2

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★GiveN:

• A circle has radius equal to 3 units and its centre lies on the line y = x – 1.

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★To FinD:

• Find the equation of the circle if it passes through

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★SolutioN:

$(x-h)² +(y−k)²=r2$

Here (h,k)→ centre

★Equation of line through centre

k=h−1−−−−(1)

& (x,y) point on circle

So, given (x,y)=(7,3)

$So, (7−h)2+(3−h+1)²=3²$

★Using equation

We get,

$28+h²−11h=0$

h=7 oR 4

k=6 oR 3

★Equation of circle corresponding (7,6)

$x²+y² −14x−12y+76=0$

Equation of circle corresponding (4,3)

$x²+y² −8x−6y+16=0$

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★AdditionaL informatioN

Formulas Related to Circles

The Circle Formulas are expressed as,

• Diameter of a Circle D = 2 × r
• Circumference of a Circle C = 2 × π × r
• Area of a Circle A = π × r2

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