Question

from any point p on the circle X2+y2-6x-8y+15=0 tangents PA,PB are drawn to the circle X2+y2-6x-8y+20=0 and to the locus of the orthocentre of the triangle PAB is a circle then​

Answer 1

The value of (a+b-c) is 6

Given,

$x^{2} +y^{2} -6x-8y+15=0$

$x^{2} +y^{2} -6x-8y+20=0$

To Find,

We have To Find the value of (a+b-c)

Solution,

Let us consider,

$S_{1} = x^{2} +y^{2} -6x-8y+15=0$........(i)

compare this equation (1) with the general equation of Circle is :

$x^{2} +y^{2} +2gx+2fy+c= 0$

so, from that we get,

2g=-6

or, g=-3

and, 2f=-8

or, f=-4

so, Center Coordinate of the circle is = (-g,-f) = (3,4)

and, Circle radius (R)=

$\sqrt{g^{2} +f^{2} +c} = \sqrt{9+16+15} \\</p><p>or, R = \sqrt{40} = 2\sqrt{10}$

similarly,

$S_{2} = x^{2} +y^{2} -6x-8y+20=0$.........(ii)

compare this equation (1) with the general equation of Circle is :

$x^{2} +y^{2} +2gx+2fy+c= 0$

2g= -6

or, g=-3

and, 2f= -8

or, f= 4

so, Center Coordinate of the circle is = (-g,-f) = (3,4)

and radius (r)= $r= \sqrt{g^{2} +f^{2} +c} = \sqrt{9+16+20} \\</p><p>or, = \sqrt{45} = 3\sqrt{5}$

Now,

(r-R) = $\sqrt{45-40} = \sqrt{5}$

and, AB= $\sqrt{16+16} =\sqrt{32}$

BD= $\frac{\sqrt{32} }{16} = \frac{\sqrt{32}*\sqrt{32} }{16*\sqrt{32} } \\= \frac{32}{16*\sqrt{32} } = \frac{2}{32}$

so, locus of the orthocentre of the ∆PAB is a circle = $x^{2} +y^{2} -6x-8y+20=0$

compare this with normal form,

$x^{2} +y^{2} + ac+ by+c=0$

so, a+b-c= -6-8+20 = 6

Hence, the value of (a+b-c) is 6

#SPJ2

Answer 2

$</p><p>S _{1} \: : \: {x}^{2} + {y}^{2} - 6x - 8y + 15 = 0 \\ ⇒( - g, - f) = (3,4) \\ r = \sqrt{ {3}^{2} + {4}^{2} - 15 } \\ ⇒r = \sqrt{10}$

$S _{2} \: : \: {x}^{2} + {y}^{2} - 6x - 8y + 20 = 0 \\ ⇒( - g, - f) = (3,4) \\ R = \sqrt{ {3}^{2} + {4}^{2} - 20 } \\ ⇒R = \sqrt{5}$

$△POB \: ⇒ \\ PO = \sqrt{10} \\ AO = OB = 3 \\ ∠PBO = 90° \\ {PO}^{2} = {OB}^{2} + {PB}^{2} \\ ⇒{( \sqrt{10} )}^{2} = {( \sqrt{5} )}^{2} + {PB}^{2} \\⇒ PB = \sqrt{5} = PA$

$△PAB \: ⇒ \\ ∠APB = 90° \\ {AB}^{2} = {AP}^{2} + {PB}^{2} \\ ⇒{(AB)}^{2} = {( \sqrt{5} )}^{2} + {( \sqrt{5} )}^{2} \\⇒ AB = \sqrt{10}$

Orthocentre of a right angled triangle is at its right angle

⇒ The locus of △PAB is S₁, (i.e)

${x}^{2} + {y}^{2} - 6x - 8y + 15 = 0$

∴ (-g,-f) = (3,4)

radius = √10