Q. Consider the circle `x^2 + y^2 - 10x - 6y + 30=0` . Let O be the centre of the circle and tangent at A(7,3) and B(5,1) meet at C. Let S=0 rpresent the family of circles passing through A and B:-

A) Area of quadrilateral OACB=4

B) Area of quadrilateral OACB=6

C)The co-ordinates of point C are (7,1)

D)The smallest possible circle of the family S=0 is `x^2 + y^2 -12x - 4y + 38=0`

### 1 Answer | Add Yours

Given equation of the circle is `x^2+y^2-10x-6y+30=0`

Rearranging, `x^2-10x+25+y^2-6y+9=-30+25+9`

`rArr (x-5)^2+(y-3)^2=2^2`

Therefore, coordinates of O are (5,3) and radius of the circle=2

Again `(x-5)^2+(y-3)^2=2^2`

Differentiating with respect to x,

`2(x-5)+2(y-3)*y'=0`

`rArr y'=(5-x)/(y-3)`

`y'(7,3)=(5-7)/(3-3)=oo`

Equation of tangent at (7,3),

`y-3=oo(x-7)`

`rArr (x-7)=0`

`rArr x=7`

Similarly, `y'(5,1)=(5-5)/(1-3)=0`

The equation of tangent at (5,1),

`y-1=0(x-5)`

`rArr y=1`

Therefore, coordinates of C are (7,1)

Considering quadrilateral OACB, side `BC=sqrt(2^2+0^2)=2`

and, side `AC=sqrt(0^2+2^2)=2`

So, it is a square with side=2, thus, `area= 2^2=4`

Equation of AB is y=x-4, or, `x-y=4`

This is the radical axis.

Equation of smallest circle passing through the points (7,3) and (5,1) is:

`(x-7)(x-5)+(y-3)(y-1)=0`

`rArr x^2+y^2-12x-4y+38=0`

Hence, Options A), C) and D) are the correct answers.

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes