Question -
Answer -
(i)┬аwhich touches both theaxes at a distance of 6 units from the origin.
A circle┬аtouchesthe axes at the points (┬▒6, 0) and (0, ┬▒6).
So, a circle has acentre (┬▒6, ┬▒6) and passes through the point (0, 6).
We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.
So, the equation is (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Where, p = 6, q = 6
(x тАУ 6)2┬а+(y тАУ 6)2┬а= r2┬атАж. (1)
Equation (1) passesthrough (0, 6)
So, (0 тАУ 6)2┬а+(6 тАУ 6)2┬а= r2
36 + 0 = r2
r = тИЪ36
= 6
We know that theequation of the circle with centre (p, q) and having radius тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Now by substitutingthe values in the equation, we get
(x ┬▒ 6)2┬а+(y ┬▒ 6)2┬а= (6)2
x2┬а┬▒12x + 36 + y2┬а┬▒ 12y + 36 = 36
x2┬а+ y2┬а┬▒12x ┬▒ 12y + 36 = 0
тИ┤ The equation of thecircle is x2┬а+ y2┬а┬▒ 12x ┬▒ 12y + 36 = 0.
(ii)┬аWhich touches x тАУ axisat a distance of 5 from the origin and radius 6 units.
A circle┬аtouchesthe x тАУ axis at the points (┬▒5, 0).
Let us assume thecentre of the circle is (┬▒5, a).
We have a circle withcentre (5, a) and passing through the point (5, 0) and having radius 6.
We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.
So, the equation is (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Where, p = 5, q = a
(x тАУ 5)2┬а+(y тАУ a)2┬а= r2┬атАж. (1)
Equation (1) passesthrough (5, 0)
So, (5 тАУ 5)2┬а+(0 тАУ 6)2┬а= r2
0 + 36 = r2
r = тИЪ36
= 6
We have got the centreat (┬▒5, ┬▒6) and having radius 6 units.
We know that theequation of the circle with centre (p, q) and having radius тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Now by substitutingthe values in the equation, we get
(x┬а┬▒┬а5)2┬а+(y ┬▒ 6)2┬а= (6)2
x2┬а┬▒10x + 25 + y2┬а┬▒ 12y + 36 = 36
x2┬а+ y2┬а┬▒10x ┬▒ 12y + 25 = 0.
тИ┤ The equation of thecircle is x2┬а+ y2┬а┬▒ 10x ┬▒ 12y + 25 = 0.
(iii)┬аWhich touches both theaxes and passes through the point (2, 1).
Let us assume thecircle touches the x-axis at the point (a, 0) and y-axis at the point (0, a).
Then the centre of thecircle is (a, a) and radius is a.
Its equation will be(x тАУ p)2┬а+ (y тАУ q)2┬а= r2
By substituting thevalues we get
(x тАУ a)2┬а+(y тАУ a)2┬а= a2┬атАж (1)
So now, equation (1)passes through P (2, 1)
By substituting thevalues we get
(2 тАУ a)2┬а+(1 тАУ a)2┬а= a2
4 тАУ 4a + a2┬а+1 тАУ 2a + a2┬а= a2
5 тАУ 6a + a2┬а=0
(a тАУ 5) (a тАУ 1) = 0
So, a = 5 or 1
Case (i)
We have got the centreat (5, 5) and having radius 5 units.
We know that theequation of the circle with centre (p, q) and having radius тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Now by substitutingthe values in the equation we get
(x тАУ 5)2┬а+(y тАУ 5)2┬а= 52
x2┬атАУ10x + 25 + y2┬атАУ 10y + 25 = 25
x2┬а+ y2┬атАУ10x тАУ 10y + 25 = 0.
тИ┤ The equation of thecircle is x2┬а+ y2┬атАУ 10x тАУ 10y + 25 = 0.
Case (ii)
We have got the centreat (1, 1) and having a radius 1 unit.
We know that theequation of the circle with centre (p, q) and having radius тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Now by substitutingthe values in the equation we get
(x тАУ 1)2┬а+(y тАУ 1)2┬а= 12
x2┬атАУ2x + 1 + y2┬атАУ 2y + 1 = 1
x2┬а+ y2┬атАУ2x тАУ 2y + 1 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬атАУ 2x тАУ 2y + 1 = 0.
(iv)┬аPassing through theorigin, radius 17 and ordinate of the centre is тАУ 15.
Let us assume theabscissa as тАШaтАЩ
We have a circle withcentre (a, тАУ 15) and passing through the point (0, 0) and having radius 17.
We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.
By using the distanceformula,
172┬а=a2┬а+ (-15)2
289 = a2┬а+225
a2┬а=64
|a| = тИЪ64
|a| = 8
a = ┬▒8 тАж. (1)
We have got the centreat (┬▒8, тАУ 15) and having radius 17 units.
We know that theequation of the circle with centre (p, q) and having radius тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Now by substitutingthe values in the equation, we get
(x ┬▒ 8)2┬а+(y тАУ 15)2┬а= 172
x2┬а┬▒16x + 64 + y2┬атАУ 30y + 225 = 289
x2┬а+ y2┬а┬▒16x тАУ 30y = 0.
тИ┤ The equation of thecircle is x2┬а+ y2┬а┬▒ 16x тАУ 30y = 0.